Spring 2026: Math 291 Daily Update

After reviewing class procedures, and items on the syllabus, we began a discussion of \(2\times 2\) matrices over \(\mathbb{R}\), the set of which we denote by \(\textrm{M}_2(\mathbb{R})\). Given \(A = \begin{pmatrix} a & b\\c & d\end{pmatrix}\), we identified the entries of \(A\) as follows: \(a\) is the (1,1) entry, \(b\) is the (1,2) entry, \(c\) is the (2,1) entry and \(d\) is the (2,2) entry.

We established two fundamental operations:

We then discussed at length the following properties. In what follows, \(A, B, C\in \textrm{M}_2(\mathbb{R})\) and \(\lambda, \lambda_1, \lambda_2 \in \mathbb{R}\).

We also discussed how one might prove these identities and noted that the properties above will be recurring throughout the semester as we discuss abstract vector spaces. We ended class by discussing the following consequences of properties (1)-(8) above. Keeping the same notation, we have

In today's lecture we introduced two new operations for \(2\times 2\) matrices, namely multiplication of a matrix times a column and multiplication of two (\(2\times 2\)) matrices.

For \(A = \begin{pmatrix} a & b\\c & d\end{pmatrix}\), \(B = \begin{pmatrix} e & f\\g & h\end{pmatrix}\), \(C = \begin{pmatrix} u\\v\end{pmatrix}\), we defined

We noted that the (1,1) entry of \(AB\) is \(R_1\cdot C_1\), the (1,2) entry is \(R_1\cdot C_2\), the (2,1) entry is \(R_2\cdot C_1\) and the (2,2) entry is \(R_2\cdot C_2\), where \(R_i\) is the \(i\)th row of \(A\) and \(C_j\) is the \(j\)th column of \(B\). We also noted that if we think of \(B\) as the matrix with columns \(C_1, C_2\), i.e., \(B = [C_1\ C_2]\), then \(AB = [AC_1\ AC_2]\), the matrix with columns \(AC_1, AC_2\).

We discussed how we can use the product of a matrix times a column to re-write a system of equations as a single matrix equation, as follows. Given the system of two equations in two unknowns

We can write this as \(AX = L\), where \(A = \begin{pmatrix} a & b\\c & d\end{pmatrix}\), \(X = \begin{pmatrix} x\\y\end{pmatrix}\) and \(L = \begin{pmatrix} u\\v\end{pmatrix}\).

We then discussed powers of \(2\times 2\) matrices and the class calculated \(A^2, A^3, A^4\) and conjectured the value of \(A^{2026}\), for \(A = \begin{pmatrix} 1 & 0\\1 & 1\end{pmatrix}\). We were easily able to conjecture \(A^n = \begin{pmatrix} 1 & 0\\n & 1\end{pmatrix}\), for all \(n\geq 1\), which led to a discussion of how to use mathematical induction to prove this fact. One first establishes the base case \(n = 1\), which in this case is clear. One then shows that the \(n-1\) case implies the \(n\)th case - the inductive step, which in this case amounted to showing that \(A\cdot \begin{pmatrix} 1 & 0\\n-1 & 1\end{pmatrix} = \begin{pmatrix} 1 & 0\\n & 1\end{pmatrix} = A^n\). The class then used induction to prove the formula \(1+2+\cdots + n = \frac{n(n+1)}{2}\).

We moved on to discuss (but not prove) the following

We finished class by noting that if the matrix equation \(AX = L\) represents a system of equations (as above) and \(A\) has an inverse, then we can multiply both sides of the matrix equation by \(A^{-1}\) to get the solution \(X = A^{-1}L\).

We began class with the following definition. For the \(2\times 2\) matrix \(A = \begin{pmatrix} a & b\\c & d\end{pmatrix}\), the determinant of \(A\), denoted \(\det A\), equals \(ad-bc\).

We then discussed and verified the following

We ended class by looking at a typical system of two linear equations in two unknowns

We noted that each equation corresponds to a straight line \(L_1, L_2\) (respectively) in \(\mathbb{R}^2\) and \((s,t)\) is a solution to the system if and only if \((s,t)\) is a point on each line. Thus the following are the only possibilities for the solution set to the given system of equations:

Thus, there can never be a \(2\times 2\) system of linear equations with exactly 17 solutions! (Or with exactly \(n\) solutions for any \(n > 1\).)

In the previous lecture we saw that given a system of linear equations

\[\begin{align*} ax+by &= u\\ cx+dy &= v \end{align*}\]

whose coefficient matrix \(A = \begin{pmatrix} a & b\\c & d\end{pmatrix}\) has non-zero determinant \(\Delta\), then the solution to the system is given by

We noted that for large systems of linear equations, Cramer's rule is not cost effective, so we began a discussion of Gaussian elimination. We started with a specific system of equations (like)

\[\begin{align*} 2x+6y &= 8\\ 3x+y &= 4 \end{align*}\]

and performed a sequence of operations that changed the system, but preserved the solution. These operations were of the following form: Interchange equations, add a multiple of one equation to another equation, and multiply an equation by a non-zero number. This simplified the system to one trivially solvable, namely

\[\begin{align*} x &= 1\\ y &= 1. \end{align*}\]

We noted that in doing the various operations, the arithmetic involved the coefficients in the equations and the variables were essentially placeholders. This led to considering the corresponding augmented matrix \(\begin{bmatrix}2 & 6 & | & 8\\3 & 1 & | & 4\end{bmatrix}\). By performing the same operations on the rows of the augmented matrix that we did on the system of equations, this led to the augmented matrix \(\begin{bmatrix}1 & 0 & | & 1\\0 & 1 & | & 1\end{bmatrix}\), which corresponds to the system

\[\begin{align*} x &= 1\\ y &= 1. \end{align*}\]

We formalized this process by defining

We noted that the goal was to put, if possible, the beginning augmented matrix \(\begin{bmatrix}a & b & | & u\\c & d & | & v\end{bmatrix}\) into the form \(\begin{bmatrix}1 & 0 & | & s\\0 & 1 & | & t\end{bmatrix}\), from which the solution \(x = s, y = t\) could be read. We noted that the strategy for the Gaussian elimination process should be as follows: Using elementary row operations, first get a 1 in the (1,1) entry of the augmented matrix, then use that 1 to get 0 below it. Then make, if possible, the (2,2) entry of the augmented matrix 1 and then use that 1 to get a 0 above it. This will always be possible when the original system has a unique solution.

We ended class by considering the remaining two cases. In one case, the final augmented matrix took the form \(\begin{bmatrix}1 & 3 & | & 4\\0 & 0 & | & 0\end{bmatrix}\), which corresponds to the system \(x+3y = 4\), from which one concludes \(x = 4-3y\). To describe the solution set we introduced another parameter \(t\) to get \(\{(4-3t, t)\ |\ t\in \mathbb{R}\}\) as the solution set. This is the case when the system has infinitely many solutions. We then saw an example where the final augmented matrix took the form \(\begin{bmatrix}1 & 3 & | & 4\\0 & 0 & | & 1\end{bmatrix}\), which meant the original system had no solution, since \(0 = 1\) is a contradiction.

After reviewing the possible outcomes for solving a system of two linear equations in two unknowns using Gaussian elimination, we considered the following \(2\times 3\) system of equations

As expected, we began with the augmented matrix \(\begin{bmatrix}2 & 5 & 2 & | & 9\\1 & 2 & -12 & | & 4\end{bmatrix}\), applied elementary row operations with the same strategy as in the \(2\times 2\) case and arrived at \(\begin{bmatrix}1 & 0 & -9 & | & 2\\0 & 1 & 4 & | & 1\end{bmatrix}\), from which we derived the solution set \(\{(2+9t,1-4t,t)\ |\ t\in \mathbb{R}\}\), which is a one parameter solution set.

We then considered the system

Following the same procedure led to the augmented matrix \(\begin{bmatrix}1 & 2 & -1 & | & 4\\0 & 0 & 0 & | & 0\end{bmatrix}\), showing that the solution set is \(\{(4-2t_1+t_2, t_1, t_2)\ |\ t_1, t_2\in \mathbb{R}\}\), a two parameter solution set. We then recorded the fact that if we start with a system of two linear equations in three unknowns, the original augmented matrix \(\begin{bmatrix}a & b & c & | & u\\d & e & f & | & v\end{bmatrix}\) takes one of the following three forms after performing elementary row operations:

where \(\alpha \neq 0\) in case C. We noted that the solution set is infinite in case A, with a one parameter solution, the solution set in Case B is infinite with a two parameter solution set and that there was no solution in Case C. In particular, no \(2\times 3\) system of linear equations has a unique solution. We further noted that in all cases we have seen, there is the

After noting that this important fact applies to systems of equations of any size, we then formalized the form the final matrix should take in the Gaussian elimination process. This applies to systems of linear equations of any size.

We ended class by noting that the technique of Gaussian elimination, namely converting the system to an augmented matrix and applying elementary row operations to get a RREF, applies to systems of linear equations of any size and we illustrated this by solving the system

which had the solution \(x = 1, y = 2, z = 3\).

We began class by reviewing the possible outcomes in terms of the RREFs for an augmented matrix representing a \(3\times 3\) system of linear equations. We then reminded the class of the important fact that given any system of linear equations, the number independent parameters needed to describe the solution set is the number of variables minus the number of leading 1s in the RREF of the corresponding augmented matrix. We also noted that a system of linear equations is said to be homogeneous if the right hand side of the system consists entirely of zeros. In this case, there will always be at least one solution, namely the solution in which all of the given variables equal zero.

We then noted the following

After computing a couple of products, we noted that the product rules (i)-(v) from January 22 hold, as long as the product exists. We then began a discussion of elementary \(2\times 2\) matrices.

The class then verified (in the \(2\times 2\) case) that for any elementary matrix \(E\), \(EA\) is the same matrix obtained by applying the corresponding elementary row operation to \(A\). We then noted that elementary matrices are invertible, and their inverses are elementary matrices that can be easily guessed.

We then asked what does it mean in terms of elementary matrices if \(A\) is a \(2\times 2\) matrix that reduces to \(I_2\) via elementary row operations. We inferred that there exists a \(2\times 2\) matrix \(H\) such that \(HA = I_2\) and thus (via the bonus problem listed in today's homework) \(A\) is invertible and \(A^{-1}\) can be found via Gaussian elimination, as stated below.

We began class by thinking of \(\mathbb{R}^2\) as a vector space even though we have not yet formally defined the general concept of a vector space. We noted that the elements of \(\mathbb{R}^2\) can be thought of as row or column vectors and as such one can add two vectors or scalar multiply a vector by a real number exactly as one does in Calculus II. We then noted that the elements of \(\mathbb{R}^2\) satisfy the Fundamental Properties from the lecture of January 20. We then pointed out (informally) that it is precisely these properties that make any set with addition and scalar multiplication into a vector space. We then gave the following

We noted that geometrically, condition (ii) just says that \(v_1\) and \(v_2\) do not lie on the same line through the origin in \(\mathbb{R}^2\). We also noted that if one of \(v_1, v_2\) is \(\vec{0}\), then \(v_1, v_2\) cannot be linearly independent. We also defined \(v_1, v_2\) to be linearly dependent if they are not linearly independent. We noted that the vectors \((1,2), (2,1)\) are linearly independent, while the vectors \((1,2), (4,8)\) are linearly dependent.

We then proved the following important theorem.

We ended class by noting, but not verifying, that if \(v_1, v_2 \in \mathbb{R}^2\) are linearly independent, then given any vector \(u\in \mathbb{R}^2\), we can find (unique) \(\alpha, \beta \in \mathbb{R}\) such that \(u = \alpha v_1+\beta v_2\).

We began class by reviewing the definition of what it means for two vectors in \(\mathbb{R}^2\) to be linearly independent. This led to a discussion and example of how any vector in \(\mathbb{R}^2\) is a linear combination of a fixed set of linearly independent vectors. In particular, we showed how \((6,9)\) can be written as a linear combination of the independent vectors \((2,3)\) and \((1,3)\). We then formally gave the following definitions.

We then established the following theorem:

We ended class by observing that a line through the origin in \(\mathbb{R}^2\) is closed under vector addition and scalar multiplication, and noted that these properties will determine the concept of subspace to be discussed next week.

We began class with the following important definition.

Examples of linear transformations given in class were:

1. (i) \(T(x,y) = (2x+2y, -6x+2y)\);
2. (ii) \(T(x,y) = (-y,x)\), rotation 90 degrees counterclockwise;
3. (iii) \(T(x,y) = (y,-x+2y)\).

We noted that the second linear transformation has the property that no line through the origin is mapped to itself by the linear transformation (i.e., \(T\) has no eigenvectors), while the third linear transformation maps the line \(y = x\) to itself (i.e., the vector \((1,1)\) is an eigenvector).

We then noted that a linear transformation \(T: \mathbb{R}^2\to \mathbb{R}^2\) is totally determined by its effect on the standard basis of \(\mathbb{R}^2\) given by \(E = \{e_1, e_2\}\), where \(e_1 = (1,0)\) and \(e_2 = (0,1)\). In fact, if \(v = (a,b)\), then

showing that the value of \(T(v)\) is already determined by the values of \(T(e_1)\) and \(T(e_2)\).

We then gave the following definition, which we noted is a special case of a more general definition to come. We first reminded the class of the theorem from the previous lecture which showed that \(v_1, v_2\in \mathbb{R}^2\) are linearly independent if and only if they span \(\mathbb{R}^2\).

After noting that the standard basis for \(\mathbb{R}^2\) mentioned above is a basis for \(\mathbb{R}^2\), we gave the following important definition.

We then noted that \([T]^E_E\) is the easiest matrix to calculate, namely if \(T(x,y) = (ax+cy, bx+dy)\), then \([T]_E^E = \begin{pmatrix} a & c\\b & d\end{pmatrix}\). From this we noted that if \(v = (e,f)\), then in terms of columns, \(T(v) = T(e,f) = A\begin{pmatrix} e\\f\end{pmatrix}\), where \(A = [T]_E^E\). This equation always holds true as long as we take the standard basis for both the input basis and output basis.

We finished class by noting that if \(T(x,y) = (2x-4y, x+7y)\), then \([T]_E^E = \begin{pmatrix} 2 & -4\\1 & 7\end{pmatrix}\), while if \(B = \{w_1, w_2\}\) with \(w_1 = (1,1)\) and \(w_2 = (1,2)\), \([T]_B^E = \begin{pmatrix} -2 & -6\\8 & 15\end{pmatrix}\), which follows easily from the fact that \(T(1,1) = (-2,8)\) and \(T(1,2) = (-6,15)\). However, to calculate \([T]_E^B\), we saw that this required solving two systems of equations, namely

After recalling the definition of a basis for \(\mathbb{R}^2\), linear transformation and the matrix of a linear transformation with respect to two bases, we considered the following question: Suppose we are given \(S,T: \mathbb{R}^2\to \mathbb{R}^2\) are two linear transformations. Then the composition \(ST:\mathbb{R}^2\to \mathbb{R}^2\) is also a linear transformation. Can a matrix of \(ST\) be expressed in terms of matrices for \(S\) and \(T\)?

After proving this formula, we described change of basis matrices, by noting that if \(\alpha, \beta\) are bases for \(\mathbb{R}^2\), then \([I_2]_{\alpha}^{\beta}\) is the matrix obtained by expressing the basis elements of \(\alpha\) in terms of the basis \(\beta\). Similarly, \([I_2]_{\beta}^{\alpha}\) is the matrix obtained by expressing the basis elements of \(\beta\) in terms of \(\alpha\). Using the very important formula above we saw that the matrices are inverses of one another:

This led to the fundamental formula, which follows from the very important formula.

The class worked on practice problems for Exam I.

Exam 1.

After making a few comments about Exam 1, we looked at the following

We approached this example by setting \(v = \begin{pmatrix} x\\y\end{pmatrix}\) and setting up a system of equations, that ultimately became the matrix equation \(\begin{pmatrix} -\lambda & -2\\1 & -\lambda+3\end{pmatrix} \begin{pmatrix} x\\y\end{pmatrix} = \begin{pmatrix} 0\\0\end{pmatrix}\). Since the vector \(v\) should be non-zero, the matrix equation should have a non-trivial solution, which happens when \(\det \begin{pmatrix} -\lambda & -2\\1 & -\lambda+3\end{pmatrix} = 0\). This led to the polynomial equation \(\lambda^2-3\lambda+2 = 0\), which has solutions \(\lambda = 2\) and \(\lambda = 1\). These are the values of \(\lambda\) we seek. What about the corresponding vectors? We saw that when \(\lambda = 2\), the matrix equation becomes \(\begin{pmatrix} -2 & -2\\1 & 1\end{pmatrix} \begin{pmatrix} x\\y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}\), which is easily seen to have a non-trivial solution \(v_1 = \begin{pmatrix} 1\\-1\end{pmatrix}\). Similarly, we saw that when \(\lambda = 1\), the matrix equation has coefficient matrix \(\begin{pmatrix} -1 & -2\\1 & 2\end{pmatrix}\), so \(v_2 = \begin{pmatrix} 2\\-1\end{pmatrix}\) is a non-trivial solution. Thus, we found two vectors \(v_1, v_2\) such that \(Av_1 = 2\cdot v_1\) and \(Av_2 = 1\cdot v_2\). We also noted that any multiple of \(v_1\) works for \(\lambda = 2\) and any multiple of \(v_2\) works for \(\lambda = 1\).

We then noted that \(v_1, v_2\) form a basis for \(\mathbb{R}^2\), since the determinant of \(P := \begin{pmatrix} 1 & 2\\-1 & -1\end{pmatrix}\) is not zero. We then saw that \(P^{-1} = \begin{pmatrix} -1 & -2\\1 & 1\end{pmatrix}\) and that if we set \(B = \begin{pmatrix} 2 & 0\\0 & 1\end{pmatrix}\), then \(B = P^{-1}AP\). Thus, we saw

This led to the following

This led to formalizing the process of finding eigenvalues.

We ended class by repeating the steps in Example A for the matrix \(B = \begin{pmatrix} -1 & 2\\4 & -3\end{pmatrix}\).

We began class by reviewing the following for \(A \in \textrm{M}_2(\mathbb{R})\):

1. (i) \(\lambda \in \mathbb{R}\) is an eigenvalue of \(A\) if \(Av = \lambda v\), for some non-zero vector \(v\in \mathbb{R}^2\).
2. (ii) Any non-zero vector satisfying \(Av = \lambda v\) is an eigenvector of \(A\) associated to the eigenvalue \(\lambda\).
3. (iii) The eigenvalues are found by solving the equation \(\det (A-\lambda I_2) = 0\) for \(\lambda\). In other words, \(\alpha\) is an eigenvalue of \(A\) if and only if \(\alpha\) is a root of the polynomial \(p_A(x) := \det (A-xI_2)\). The polynomial \(p_A(x)\) is called the characteristic polynomial of \(A\).
4. (iv) If \(\alpha\) is an eigenvalue of \(A\), the corresponding eigenvectors are the vectors \(v\in \mathbb{R}^2\) solving the matrix equation \((A-\alpha I_2)\cdot v = \vec{0}\), or equivalently, the ordered pairs that are solutions to the system of equations \[\begin{aligned}(-\alpha +a)x+by &=0\\ cx+(-\alpha + d)y &= 0,\end{aligned}\] where \(A = \begin{pmatrix} a & b\\c & d\end{pmatrix}\).
5. (v) \(A\) is said to be diagonalizable if there exists invertible \(P\in \textrm{M}_2(\mathbb{R})\) such that \(P^{-1}AP = \begin{pmatrix} \alpha & 0\\0 & \beta\end{pmatrix}\), for some \(\alpha, \beta \in \mathbb{R}\).

We then considered the following

We then observed that the characteristic polynomial of the matrix \(A = \begin{pmatrix} 0 & -1\\1 & 0\end{pmatrix}\) is \(x^2+1\), so \(A\) does not have any (real) eigenvalues. This was explained geometrically, since multiplication of a vector in \(\mathbb{R}^2\) by \(A\) rotates the vector 90 degrees counter-clockwise, so that no vector in \(\mathbb{R}^2\) is mapped to a multiple of itself.

We then stated the following

We ended class by stating, but not proving the following theorem, which formalizes the processes used in our previous examples.

We began class by re-stating the Diagonalizability Theorem from the end of the previous lecture. With the theorem in mind, we then considered the following three scenarios.

We noted that we have previously seen examples of types (i) and (iii). We then noted that for \(A = \begin{pmatrix} 2 & 7\\0 & 2\end{pmatrix}\), \(p_A(x) = (x-2)^2\) and that all eigenvectors are multiples of \(v = \begin{pmatrix} 1\\0\end{pmatrix}\), so that by the theorem, \(A\) is not diagonalizable.

We then gave a proof of the theorem characterizing diagonalizability. The proof of the theorem relied heavily on the two Key Observations from the previous lecture. We finished class by presenting the following very important corollary to the diagonalizability theorem.

Most of the class was spent discussing how diagonalizability of a \(2\times 2\) matrix can be used to solve a coupled system of two linear first order differential equations. In particular, given the system

assuming \(A\) is diagonalizable, we worked through the derivation of the solution to the system and found that the solution took the form

where \(\alpha, \beta\) are the eigenvalues of \(A\), with corresponding eigenvectors \(v_1, v_2\in \mathbb{R}^2\), and where \(c_1, c_2\in \mathbb{R}\) are determined by the initial conditions \(\begin{pmatrix} x_1(0)\\x_2(0)\end{pmatrix} = P\cdot \begin{pmatrix} c_1\\c_2\end{pmatrix}\), for \(P = [v_1\ v_2]\).

We finished class by using the derivation above to solve the system

with initial conditions \(x_1(0) = 3, x_2(0) = -4\).

We began class by discussing our second application of diagonalizability, namely, that if \(A\) is diagonalizable, it is easy to write down a formula for the \(n\)th power of \(A\), as follows, using the fact that for any square matrix \(C\), \((PCP^{-1})^n = PC^nP^{-1}\): If \(P^{-1}AP = D = \begin{pmatrix} \alpha & 0\\0 & \beta \end{pmatrix}\), then \(A = PDP^{-1}\) and

We then calculated \(A^n\) for \(A = \begin{pmatrix} -6 & 10\\-5 & 9\end{pmatrix}\) and \(n\geq 1\), and found that \(A^n = \begin{pmatrix} 2(-1)^n - 4^n & 2(-1)^{n+1}+2\cdot 4^n\\(-1)^{n}-4^n & (-1)^{n+1}+2\cdot 4^n\end{pmatrix}\).

We then defined \(e^A\), for \(A\) a diagonalizable matrix, by substituting \(A\) into the partial sum \(1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots+\frac{1}{n!}x^n\) and taking a limit in each entry. We noted, but did not prove, that each entry does converge to a real number so that \(e^A\) does exist. We noted that the entries of \(e^A\) are not \(e\) raised to the corresponding entry, but did note the important fact that if \(D = \begin{pmatrix} \alpha & 0\\0 & \beta\end{pmatrix}\), then \(e^D = \begin{pmatrix} e^{\alpha} & 0\\0 & e^{\beta}\end{pmatrix}\). When \(A\) is diagonalizable and \(P^{-1}AP = D\), putting \(A = PDP^{-1}\) into the partial sum expression enabled us to see that \(e^A = Pe^DP^{-1}\), which can easily be calculated.

We then calculated \(e^A\), for \(A = \begin{pmatrix} -6 & 10\\-5 & 9\end{pmatrix}\) and got \(e^A = \begin{pmatrix} 2e^{-1}-e^4 & -2e^{-1}+2e^4\\e^{-1}-e^4 & -e^{-1}+2e^4\end{pmatrix}\).

We then reconciled our two applications by noting that if

is a coupled system of first order linear differential equations that can be written as \(X'(t) = A\cdot X(t)\), with \(A = \begin{pmatrix} a & b\\c & d\end{pmatrix}\), then the solution takes the very agreeable form \(X(t) = e^{At}\cdot \begin{pmatrix} x_1(0)\\x_2(0)\end{pmatrix}\), which almost looks like the solution in the simple case of one function and one equation.

We ended class with yet a third application of diagonalizability, namely calculating the general term in the Fibonacci sequence: \(1, 1, 2, 3, 5, 8, 13, \ldots\), where \(a_0 = 1, a_1 = 1, a_2 = 2, a_3 = 3, \ldots\), and \(a_{k+2} = a_{k+1}+a_k\), for \(k\geq 0\). We noted that \(A^k\cdot v_0 = \begin{pmatrix} a_k\\a_{k+1}\end{pmatrix}\), for \(A = \begin{pmatrix} 0 & 1\\1 & 1\end{pmatrix}\) and \(v_0 = \begin{pmatrix} 0\\1\end{pmatrix}\). We saw that \(A\) was diagonalizable with eigenvalues \(\lambda_1 = \frac{1+\sqrt{5}}{2}\), \(\lambda_2 = \frac{1-\sqrt{5}}{2}\), so that we were able to calculate \(A^k\). This led to the closed form formula for the \(k\)th Fibonacci number as

which is an integer!

We began today's lecture by considering the following:

We then considered the symmetric matrix \(A = \begin{pmatrix} 2 & 1\\1 & 2\end{pmatrix}\), which had eigenvalues 1, 3 with associated eigenvectors \(v_1 = \begin{pmatrix} 1\\-1\end{pmatrix}\) and \(v_2 = \begin{pmatrix} 1\\1\end{pmatrix}\) respectively. We noted the crucial fact that \(v_1\) and \(v_2\) are orthogonal. We then noted that if we take \(u_1 := \frac{1}{\sqrt{2}}v_1\) and \(u_2 = \frac{1}{\sqrt{2}}v_2\), we now have two orthogonal vectors of unit length. In this case we call \(\{u_1, u_2\}\) an orthonormal basis for \(\mathbb{R}^2\). We then noted that if \(Q = [u_1\ u_2]\), then \(Q^{-1}AQ = \begin{pmatrix} 1 & 0\\0 & 3\end{pmatrix}\), so that \(A\) is orthogonally diagonalizable. This led to the following:

We then worked through the details showing that the symmetric matrix \(B = \begin{pmatrix} 2 & -2\\-2 & -1\end{pmatrix}\) is orthogonally diagonalizable. We were then able to state the important theorem.

We finished class by using the quadratic formula to work through the following proposition which gives us crucial information about \(2\times 2\) real symmetric matrices.

We immediately noted that Proposition A tells us that any real \(2\times 2\) symmetric matrix is diagonalizable (by the Corollary from the lecture of March 10).

We continued our discussion of symmetric \(2\times 2\) matrices over \(\mathbb{R}\) by restating the Spectral Theorem from the previous lecture and recalling the terminology in the statement of the theorem. We also recalled that Proposition A from the previous lecture guarantees that a real \(2\times 2\) symmetric matrix that is not a scalar matrix has two distinct eigenvalues and is thus diagonalizable. Thus the key to orthogonal diagonalizability rests with the fact that the associated eigenvectors are orthogonal. For this we needed

We had each half of the class calculate one side of the equation, and two students presented their calculation at the board. With this property in hand we were able to establish the following crucial fact.

We were then able to give

We began class by recalling that if \(A\in \mathrm{M}_2(\mathbb{R})\), then we have three possibilities for \(p_A(x)\): Either \(p_A(x)\) has two distinct real roots, \(p_A(x)\) has a repeated real root, or \(p_A(x)\) does not have a real root. We noted that we have seen many times that in the first case, \(A\) is diagonalizable, while in the second case, if \(A\) is not a scalar matrix, then if \(\lambda\) is the repeated root of \(p_A(x)\), then the eigenspace of \(\lambda\), namely, the nullspace of \(A-\lambda I_2\), is one dimensional, and thus \(A\) is not diagonalizable.

We then analyzed in depth the example \(A = \begin{pmatrix} 1 & -2\\2 & 5\end{pmatrix}\). In this case \(p_A(x) = (x-3)^2\). We confirmed \(A\) is not diagonalizable by noting that the solution space of the homogeneous system with coefficient matrix \(\begin{pmatrix} 1-3 & -2\\2 & 5-3\end{pmatrix}\) is spanned by the vector \(w := \begin{pmatrix} 1\\-1\end{pmatrix}\). We took \(v_2 = \begin{pmatrix} 1\\0\end{pmatrix}\), which is not an eigenvector. We then set \(v_1 := (A-3I_2)v_2\) to get \(v_1 = \begin{pmatrix} -2\\2\end{pmatrix}\), which is an eigenvector. Taking \(P = [v_1\ v_2] = \begin{pmatrix} -2 & 1\\2 & 0\end{pmatrix}\) we saw that \(P^{-1}AP = \begin{pmatrix} 3 & 1\\0 & 3\end{pmatrix}\). We noted that this latter matrix is the Jordan canonical form of \(A\).

To explain why this example works, we noted that from the definition of \(v_1\), we have \(v_1 = Av_2-3v_2\), so that \(Av_2 = v_1+3v_2\). Thus,

which gives \(P^{-1}AP = \begin{pmatrix} 3 & 1\\0 & 3\end{pmatrix}\).

We then stated the following

We ended class by briefly discussing what we will do in the three remaining classes before the second midterm exam on April 16.

Today we spent most of the class discussing properties of the complex numbers \(\mathbb{C}\) with the goal of noting that almost all of the linear algebra we have been doing can be done in exactly the same manner over \(\mathbb{C}\) as our previous efforts over \(\mathbb{R}\). We then discussed the following properties of complex numbers.

We then stated the Very Important

We noted the importance of this fact for linear algebra: The characteristic polynomial of any square matrix over \(\mathbb{R}\) has all of its roots in \(\mathbb{C}\), i.e., any such matrix has all of its eigenvalues in \(\mathbb{C}\). We then showed that any complex number has two square roots, meaning that we are free to solve degree two polynomials with complex coefficients using the quadratic formula.

We ended class by noting that almost all of the linear algebra we have done thus far, e.g., solving equations using Gaussian elimination, finding eigenvalues and eigenvectors, diagonalizing matrices, finding JCF, etc., can be done in exactly the same way over \(\mathbb{C}\) as we have been doing over \(\mathbb{R}\), with a minor difference arising from orthogonal diagonalization. We noted that over \(\mathbb{C}\) we have to alter the dot product of two vectors as follows: Suppose \(v_1 = \begin{pmatrix} \alpha\\\beta\end{pmatrix}\) and \(v_2 = \begin{pmatrix} \gamma\\\delta\end{pmatrix}\) with \(\alpha, \beta, \gamma, \delta\in \mathbb{C}\), then \(v_1\cdot v_2 := \alpha\overline{\gamma}+\beta\overline{\delta}\). This definition ensures that over \(\mathbb{C}\) the dot product of a non-zero vector with itself is non-zero.

We began class by pointing out that virtually everything we have done this semester works in exactly the same way over \(\mathbb{C}\) as over \(\mathbb{R}\), except for issues related to orthogonality, since the dot product over \(\mathbb{C}\) involves complex conjugation. We illustrated this principle by working through the following examples.

The class worked in groups on practice problems for Exam 2.

Exam 2.

We began a review of topics already covered, but in a general setting. Thus, we discussed the operations of sum and scalar multiplication for \(m\times n\) matrices and the usual properties that hold regarding these operations; the product of an \(m\times n\) matrix with an \(n\times p\) matrix and the usual arithmetic properties that hold; invertibility of \(n\times n\) matrices; and determinants of \(n\times n\) matrices. Regarding determinants, we first showed how to calculate the determinant of a \(3\times 3\) matrix by expanding along the first row or second column. Then we gave the following formulas:

We ended class by discussing the following properties — previously discussed for \(2\times 2\) matrices — noting that these can be used to simplify determinant calculations.

We continued with revisiting concepts we have covered throughout the semester, but from a more general point of view, beginning with a corollary to item (vi) of the properties of determinants given in the previous lecture:

We then discussed the concepts of linear dependence and linear independence for column vectors in \(\mathbb{R}^n\):

We pointed out the important facts that any set of vectors spanning \(\mathbb{R}^n\) must have at least \(n\) elements and any set of linearly independent vectors can have no more than \(n\) elements. Thus, a basis for \(\mathbb{R}^n\) always has \(n\) elements, where a basis is a collection of linearly independent vectors that span \(\mathbb{R}^n\).

We again stated the following key observation, in our general setting.

As an immediate corollary we have the following fundamental theorem.

We then discussed how the process of solving a system of \(m\) equations in \(n\) unknowns is exactly the same as for solving smaller systems of equations: one applies elementary row operations to the corresponding augmented matrix until the left hand side is in reduced row echelon form, where one can just read off the solutions. We ended class by writing down the solution set to a system of equations whose augmented matrix in reduced row echelon form was:

Using the variables \(x_1, \ldots, x_6\), this augmented matrix gives

Using the independent parameters \(t_1, t_2\), we saw that the solution set was given by

We finished by observing that if the same coefficient matrix was used for a homogeneous system of equations, then the vectors \(\begin{pmatrix} -5\\-4\\1\\0\\0\\0\end{pmatrix}\), \(\begin{pmatrix} -2\\1\\0\\-5\\1\\0\end{pmatrix}\) form a basis for the solution space.

We continued with revisiting concepts we have covered throughout the semester, but from a more general point of view, beginning with discussing the following:

We then recalled that a collection of vectors in \(\mathbb{R}^n\) (or \(\mathbb{C}^n\)) is a basis if the vectors are both linearly independent and span \(\mathbb{R}^n\). We noted that the Exchange Theorem implies that all bases for \(\mathbb{R}^n\) (or \(\mathbb{C}^n\)) have \(n\) elements. We then stated the following general version of a theorem we have used throughout the semester for vectors in \(\mathbb{R}^2\) or \(\mathbb{R}^3\).

We then began a discussion of eigenvalues and diagonalizability. Working over \(\mathbb{R}\) or \(\mathbb{C}\) we considered an \(n\times n\) matrix \(A\) with eigenvalue \(\lambda\). Writing \(p_A(x) = (x-\lambda)^e g(x)\), with \(g(\lambda) \not= 0\), we defined \(e\) to be the algebraic multiplicity of \(\lambda\). We defined \(E_\lambda\), the eigenspace of \(\lambda\), to be the solution space of the homogeneous system of equations with \(A-\lambda I_n\) as its coefficient matrix. Only the eigenspace terminology is new, as we have been studying this solution space for several weeks. Then we defined the geometric multiplicity of \(\lambda\) to be the number of independent parameters needed to describe the solutions in the eigenspace \(E_\lambda\).

We then stated the following important theorem.

We gave a proof of this theorem in the special case that \(A\) is a \(3\times 3\) matrix over \(\mathbb{R}\) with eigenvalue \(\lambda\) such that \(\lambda\) has geometric multiplicity equal to 2.

We ended class by stating one of the most important theorems for our course, which has been illustrated by various examples we have looked at during all of our recent discussions involving diagonalizability and non-diagonalizability.

We began class by re-stating and discussing the Diagonalizability Theorem presented at the end of the previous lecture. We noted that the proof of the theorem (which we did not give) depends on the following

Using the theorem we were able to derive

We then presented the following special case of the crucial (iii) implies (i) in the Diagonalizability Theorem.

We noted that the key to the proof was that the existence of two independent vectors in \(E_{\lambda_1}\) enabled us to create the invertible matrix \(P\). If \(E_{\lambda_1}\) had only one independent eigenvector, we could not have created a \(3\times 3\) matrix of eigenvectors with linearly independent columns.

We then ended the discussion of general statements of topics covered this semester by stating

We then began a discussion of vector spaces in the abstract, starting with the definition below. We agreed to write \(F\) to indicate either \(\mathbb{R}\) or \(\mathbb{C}\).

We ended class by giving the following examples and discussing how they are all similar, in terms of their underlying vector space structure.