Fall 2026: Math 590 Exam 1 Practice Problems with Solutions

Problem 1

For each of the following subsets, determine whether or not it is a subspace of the given vector space. For those that are subspaces, find a basis and give the dimension.

(a) $W = \left\{\begin{pmatrix} x\\y\\z\end{pmatrix}\in\mathbb{R}^3\ \Big|\ x+2y-3z = 0\right\}$.
(b) $W = \left\{\begin{pmatrix} x\\y\\z\end{pmatrix}\in\mathbb{R}^3\ \Big|\ xy = 0\right\}$.
(c) $W = \left\{A\in\mathrm{M}_2(\mathbb{R})\ \Big|\ A\begin{pmatrix} 1\\1\end{pmatrix} = \begin{pmatrix} 0\\0\end{pmatrix}\right\}$.
(d) $W = \{p(x)\in P(2)\ |\ p(1) = 2p(0)\}$.
(e) $W = \{p(x)\in P(2)\ |\ p(0) = 1\}$.

Solution. (a) $W$ is a subspace, since it is the solution space of a homogeneous system of linear equations (a single equation in three unknowns), and such solution spaces are subspaces, as noted in the lecture of August 26. Solving $x = -2y+3z$, with $y, z$ free, every element of $W$ has the form

\begin{pmatrix} -2y+3z\\y\\z\end{pmatrix} = y\begin{pmatrix} -2\\1\\0\end{pmatrix}+z\begin{pmatrix} 3\\0\\1\end{pmatrix}.

Thus $\left\{\begin{pmatrix} -2\\1\\0\end{pmatrix}, \begin{pmatrix} 3\\0\\1\end{pmatrix}\right\}$ spans $W$, and these vectors are linearly independent (look at the second and third coordinates), so they form a basis for $W$ and $\dim(W) = 2$.

(b) $W$ is not a subspace, since it is not closed under addition, condition (i) in the definition of subspace given in the lecture of August 26: $\begin{pmatrix} 1\\0\\0\end{pmatrix}, \begin{pmatrix} 0\\1\\0\end{pmatrix}\in W$, but their sum $\begin{pmatrix} 1\\1\\0\end{pmatrix}\not\in W$, since $1\cdot 1 = 1\not= 0$.

(c) $W$ is a subspace. Write $u = \begin{pmatrix} 1\\1\end{pmatrix}$. If $A, B\in W$ and $\lambda\in\mathbb{R}$, then $(A+B)u = Au+Bu = \vec{0}+\vec{0} = \vec{0}$ and $(\lambda A)u = \lambda(Au) = \lambda\vec{0} = \vec{0}$, so $A+B, \lambda A\in W$. For $A = \begin{pmatrix} a & b\\c & d\end{pmatrix}$, $Au = \begin{pmatrix} a+b\\c+d\end{pmatrix}$, so $A\in W$ if and only if $b = -a$ and $d = -c$, i.e.,

A = \begin{pmatrix} a & -a\\c & -c\end{pmatrix} = a\begin{pmatrix} 1 & -1\\0 & 0\end{pmatrix}+c\begin{pmatrix} 0 & 0\\1 & -1\end{pmatrix}.

The two matrices on the right span $W$ and are clearly linearly independent, so they form a basis for $W$ and $\dim(W) = 2$.

(d) $W$ is a subspace: if $p(1) = 2p(0)$ and $q(1) = 2q(0)$, then $(p+q)(1) = p(1)+q(1) = 2p(0)+2q(0) = 2(p+q)(0)$, and $(\lambda p)(1) = \lambda p(1) = \lambda\cdot 2p(0) = 2(\lambda p)(0)$. Writing $p(x) = a+bx+cx^2$, the condition $p(1) = 2p(0)$ becomes $a+b+c = 2a$, i.e., $a = b+c$. Thus

\[p(x) = (b+c)+bx+cx^2 = b(1+x)+c(1+x^2),\]

so $1+x, 1+x^2$ span $W$. They are linearly independent (neither is a multiple of the other), so $\{1+x, 1+x^2\}$ is a basis for $W$ and $\dim(W) = 2$.

(e) $W$ is not a subspace: as noted in the lecture of August 26, every subspace contains $\vec{0}$, but the zero polynomial satisfies $0(0) = 0\not= 1$, so $\vec{0}\not\in W$. (Alternatively, $1\in W$, but $1+1 = 2\not\in W$.)

Problem 2

Let $v_1 = \begin{pmatrix} 1\\2\\0\\1\end{pmatrix}$, $v_2 = \begin{pmatrix} 0\\1\\1\\-1\end{pmatrix}$, $v_3 = \begin{pmatrix} 2\\3\\-1\\3\end{pmatrix}$ and $w = \begin{pmatrix} 3\\5\\h\\4\end{pmatrix}$ in $\mathbb{R}^4$.

(a) Find all values of $h$ for which $w\in\mathrm{Span}\{v_1, v_2, v_3\}$.
(b) Are $v_1, v_2, v_3$ linearly independent? If not, find a non-trivial dependence relation among them.
(c) For the value(s) of $h$ found in (a), write $w$ as a linear combination of $v_1, v_2, v_3$ in two different ways. Why doesn't this contradict uniqueness of representation?
(d) Find a basis for $\mathrm{Span}\{v_1, v_2, v_3\}$ and give its dimension.

Solution. (a) As discussed in the lecture of August 31 (see also the Summary in the lecture of September 2), $w\in\mathrm{Span}\{v_1, v_2, v_3\}$ if and only if the system with augmented matrix $[v_1\ v_2\ v_3\ |\ w]$ has a solution. Row reducing ($R_2-2R_1$, $R_4-R_1$, then $R_3-R_2$, $R_4+R_2$):

\begin{aligned} \left[\begin{array}{ccc|c} 1 & 0 & 2 & 3\\2 & 1 & 3 & 5\\0 & 1 & -1 & h\\1 & -1 & 3 & 4\end{array}\right] &\longrightarrow \left[\begin{array}{ccc|c} 1 & 0 & 2 & 3\\0 & 1 & -1 & -1\\0 & 1 & -1 & h\\0 & -1 & 1 & 1\end{array}\right]\\[6pt] &\longrightarrow \left[\begin{array}{ccc|c} 1 & 0 & 2 & 3\\0 & 1 & -1 & -1\\0 & 0 & 0 & h+1\\0 & 0 & 0 & 0\end{array}\right]. \end{aligned}

The system is consistent if and only if $h+1 = 0$. Thus $w\in\mathrm{Span}\{v_1, v_2, v_3\}$ if and only if $h = -1$.

(b) From the RREF of the coefficient matrix above, the homogeneous system $x_1v_1+x_2v_2+x_3v_3 = \vec{0}$ has solutions $x_1 = -2t$, $x_2 = t$, $x_3 = t$. Taking $t = -1$ gives the non-trivial dependence relation $2v_1-v_2-v_3 = \vec{0}$, i.e., $v_3 = 2v_1-v_2$. By the Observation in the lecture of September 2, $v_1, v_2, v_3$ are linearly dependent.

(c) For $h = -1$, the solutions of the system in (a) are $x_1 = 3-2t$, $x_2 = -1+t$, $x_3 = t$. Taking $t = 0$ and $t = 1$ gives

w = 3v_1-v_2+0\cdot v_3 \quad \text{and}\quad w = v_1+0\cdot v_2+v_3.

(Check: $v_1+v_3 = \begin{pmatrix} 3\\5\\-1\\4\end{pmatrix} = w$.) This does not contradict Uniqueness of Representation, from the lecture of September 11, since that result requires $v_1, v_2, v_3$ to be a basis, and by (b) they are not linearly independent.

(d) The pivots in the RREF of $[v_1\ v_2\ v_3]$ lie in columns 1 and 2, so by the algorithm for finding a basis, given in the lecture of September 14, $\{v_1, v_2\}$ is a basis for $\mathrm{Span}\{v_1, v_2, v_3\}$, which therefore has dimension 2.

Problem 3

For $k\in\mathbb{C}$, consider the vectors

v_1 = \begin{pmatrix} 1\\k\\0\end{pmatrix},\quad v_2 = \begin{pmatrix} 0\\1\\k\end{pmatrix},\quad v_3 = \begin{pmatrix} k\\0\\1\end{pmatrix}

in $\mathbb{C}^3$.

(a) Find all $k\in\mathbb{C}$ for which $v_1, v_2, v_3$ are linearly dependent.
(b) For each real value of $k$ found in (a), find a non-trivial dependence relation among $v_1, v_2, v_3$.
(c) For $k = 1$, explain without further calculation why $v_1, v_2, v_3$ form a basis for $\mathbb{C}^3$, and why the system $AX = b$ has a unique solution for every $b\in\mathbb{C}^3$, where $A = [v_1\ v_2\ v_3]$.

Solution. (a) Let $A = [v_1\ v_2\ v_3] = \begin{pmatrix} 1 & 0 & k\\k & 1 & 0\\0 & k & 1\end{pmatrix}$. By the Very Important Special Case of Column Vectors, given in the lecture of September 11, the vectors are linearly dependent if and only if $|A| = 0$. Expanding along the first row,

|A| = 1\cdot\begin{vmatrix} 1 & 0\\k & 1\end{vmatrix}-0+k\cdot\begin{vmatrix} k & 1\\0 & k\end{vmatrix} = 1+k^3 = (k+1)(k^2-k+1).

Thus $v_1, v_2, v_3$ are linearly dependent if and only if $k^3 = -1$, i.e.,

k = -1,\quad k = \frac{1+i\sqrt{3}}{2},\quad k = \frac{1-i\sqrt{3}}{2}.

(b) The only real value is $k = -1$, where $v_1 = \begin{pmatrix} 1\\-1\\0\end{pmatrix}$, $v_2 = \begin{pmatrix} 0\\1\\-1\end{pmatrix}$, $v_3 = \begin{pmatrix} -1\\0\\1\end{pmatrix}$. Then $v_1+v_2+v_3 = \vec{0}$, a non-trivial dependence relation.

(c) For $k = 1$, $|A| = 1+1 = 2\not= 0$. By the theorem stated in the lecture of September 25 (see also the Very Important Special Case of Column Vectors from the lecture of September 11), the columns $v_1, v_2, v_3$ of $A$ are linearly independent and span $\mathbb{C}^3$, so they form a basis for $\mathbb{C}^3$; and $A$ is invertible, so $AX = b$ has the unique solution $X = A^{-1}b$ for every $b\in\mathbb{C}^3$.

Problem 4

Let

A = \begin{pmatrix} 1 & 2 & 0 & 1 & 1\\2 & 4 & 1 & 3 & 0\\-1 & -2 & 1 & 0 & 1\\0 & 0 & 1 & 1 & 2\end{pmatrix},

and let $W\subseteq\mathbb{R}^4$ be the subspace spanned by the columns $C_1, \ldots, C_5$ of $A$.

(a) Find the RREF of $A$.
(b) Use the algorithm for finding a basis to find a basis for $W$ consisting of columns of $A$. What is $\dim(W)$?
(c) Express each column of $A$ not in your basis as a linear combination of the basis vectors.
(d) Find a basis for the solution space of the homogeneous system $AX = \vec{0}$ and give its dimension.
(e) Extend your basis for $W$ from part (b) to a basis for $\mathbb{R}^4$.

Solution. (a) Using $R_2-2R_1$, $R_3+R_1$, then $R_3-R_2$, $R_4-R_2$, and finally $\frac{1}{4}R_3$, $R_4-4R_3$, $R_2+2R_3$, $R_1-R_3$:

\begin{aligned} A &\longrightarrow \begin{pmatrix} 1 & 2 & 0 & 1 & 1\\0 & 0 & 1 & 1 & -2\\0 & 0 & 1 & 1 & 2\\0 & 0 & 1 & 1 & 2\end{pmatrix} \longrightarrow \begin{pmatrix} 1 & 2 & 0 & 1 & 1\\0 & 0 & 1 & 1 & -2\\0 & 0 & 0 & 0 & 4\\0 & 0 & 0 & 0 & 4\end{pmatrix}\\[6pt] &\longrightarrow R = \begin{pmatrix} 1 & 2 & 0 & 1 & 0\\0 & 0 & 1 & 1 & 0\\0 & 0 & 0 & 0 & 1\\0 & 0 & 0 & 0 & 0\end{pmatrix}. \end{aligned}

(b) The pivots of $R$ lie in columns 1, 3 and 5. By the algorithm for finding a basis, given in the lecture of September 14, $\{C_1, C_3, C_5\}$ is a basis for $W$, and $\dim(W) = 3$.

(c) As shown in the lecture of September 14, elementary row operations preserve solution spaces, so dependence relations among the columns of $R$ are the same as those among the columns of $A$. In $R$, column 2 is $2\cdot$(column 1) and column 4 is (column 1)$+$(column 3). Therefore

C_2 = 2C_1 \quad\text{and}\quad C_4 = C_1+C_3,

which is easily checked directly.

(d) From $R$, the solutions of $AX = \vec{0}$ satisfy $x_1 = -2x_2-x_4$, $x_3 = -x_4$, $x_5 = 0$, with $x_2, x_4$ free. Thus

X = x_2\begin{pmatrix} -2\\1\\0\\0\\0\end{pmatrix}+x_4\begin{pmatrix} -1\\0\\-1\\1\\0\end{pmatrix},

and these two vectors form a basis for the solution space, which has dimension $2 = 5-3$, in agreement with the Fundamental Fact from the lecture of August 31.

(e) Since $\dim(\mathbb{R}^4) = 4$ (Example (i) in the lecture of September 9), we need one more vector. Try $e_1$: expanding along the last column,

\begin{aligned} \big|[C_1\ C_3\ C_5\ e_1]\big| &= \begin{vmatrix} 1 & 0 & 1 & 1\\2 & 1 & 0 & 0\\-1 & 1 & 1 & 0\\0 & 1 & 2 & 0\end{vmatrix} = -\begin{vmatrix} 2 & 1 & 0\\-1 & 1 & 1\\0 & 1 & 2\end{vmatrix}\\[6pt] &= -\big(2(2-1)-1(-2-0)\big) = -4\not= 0. \end{aligned}

By the Very Important Special Case of Column Vectors from the lecture of September 11, $C_1, C_3, C_5, e_1$ are four linearly independent vectors in $\mathbb{R}^4$, and hence form a basis for $\mathbb{R}^4$. (Other choices are possible.)

Problem 5

Let $W_1 = \mathrm{Span}\left\{\begin{pmatrix} 1\\0\\1\\0\end{pmatrix}, \begin{pmatrix} 0\\1\\1\\1\end{pmatrix}\right\}$ and $W_2 = \mathrm{Span}\left\{\begin{pmatrix} 1\\1\\0\\0\end{pmatrix}, \begin{pmatrix} 1\\1\\2\\1\end{pmatrix}\right\}$ be subspaces of $\mathbb{R}^4$.

(a) Find a basis for $W_1+W_2$ and give its dimension.
(b) Find a basis for $W_1\cap W_2$ and give its dimension.
(c) Is the sum $W_1+W_2$ a direct sum? Is $\mathbb{R}^4 = W_1+W_2$?
(d) Find a subspace $W_3\subseteq\mathbb{R}^4$ such that $\mathbb{R}^4 = W_1\bigoplus W_3$.

Solution. Write $a_1 = \begin{pmatrix} 1\\0\\1\\0\end{pmatrix}$, $a_2 = \begin{pmatrix} 0\\1\\1\\1\end{pmatrix}$, $b_1 = \begin{pmatrix} 1\\1\\0\\0\end{pmatrix}$, $b_2 = \begin{pmatrix} 1\\1\\2\\1\end{pmatrix}$. Note that a typical element of $W_1$ is $aa_1+ba_2 = \begin{pmatrix} a\\b\\a+b\\b\end{pmatrix}$, so $v = \begin{pmatrix} x_1\\x_2\\x_3\\x_4\end{pmatrix}\in W_1$ if and only if $x_3 = x_1+x_2$ and $x_4 = x_2$.

(a) By the Important Fact from the lecture of August 28, $W_1+W_2 = \mathrm{Span}\{a_1, a_2, b_1, b_2\}$. Observe $b_2 = a_1+a_2$, so $W_1+W_2 = \mathrm{Span}\{a_1, a_2, b_1\}$. These three are linearly independent: $b_1\not\in W_1$, since its third coordinate $0\not= 1+1$, and $a_1, a_2$ are linearly independent. (Alternatively, the RREF of $[a_1\ a_2\ b_1\ b_2]$ has pivots in columns 1, 2, 3, and one can apply the algorithm for finding a basis from the lecture of September 14.) Thus $\{a_1, a_2, b_1\}$ is a basis for $W_1+W_2$ and $\dim(W_1+W_2) = 3$.

(b) A typical element of $W_2$ is $sb_1+tb_2 = \begin{pmatrix} s+t\\s+t\\2t\\t\end{pmatrix}$. By the description of $W_1$ above, this lies in $W_1$ if and only if $2t = (s+t)+(s+t)$ and $t = s+t$, i.e., if and only if $s = 0$. Thus $W_1\cap W_2 = \mathrm{Span}\{b_2\} = \mathrm{Span}\left\{\begin{pmatrix} 1\\1\\2\\1\end{pmatrix}\right\}$, with basis $\{b_2\}$, and $\dim(W_1\cap W_2) = 1$.

(c) The sum is not direct, since $W_1\cap W_2\not= \{\vec{0}\}$; see the definition of direct sum in the lecture of August 28. And $W_1+W_2\not= \mathbb{R}^4$, since $\dim(W_1+W_2) = 3 < 4$.

(d) Take $W_3 = \mathrm{Span}\{e_1, e_2\}$, the set of vectors of the form $\begin{pmatrix} c\\d\\0\\0\end{pmatrix}$. If $v\in W_1\cap W_3$, then $x_3 = x_4 = 0$, so $x_2 = x_4 = 0$ and $x_1 = x_3-x_2 = 0$; thus $W_1\cap W_3 = \{\vec{0}\}$. Moreover, if $aa_1+ba_2+ce_1+de_2 = \vec{0}$, then comparing coordinates gives $a+c = 0$, $b+d = 0$, $a+b = 0$, $b = 0$, so $a = b = c = d = 0$. Thus $a_1, a_2, e_1, e_2$ are four linearly independent vectors in $\mathbb{R}^4$, hence a basis by the theorem in the lecture of September 11, and by the Important Fact from the lecture of August 28, $W_1+W_3 = \mathrm{Span}\{a_1, a_2, e_1, e_2\} = \mathbb{R}^4$. Therefore, by the definition of direct sum in the lecture of August 28, $\mathbb{R}^4 = W_1\bigoplus W_3$. (Other choices of $W_3$ are possible.)

Problem 6

Let $P(3)$ denote the vector space of real polynomials of degree less than or equal to three, and let

\alpha = \{1,\ 1+x,\ (1+x)^2,\ (1+x)^3\}.

(a) Show that $\alpha$ is a basis for $P(3)$.
(b) For $p(x) = 2-x+3x^2+x^3$, find $[p(x)]_\alpha$.
(c) Let $\beta = \{1, x, x^2, x^3\}$. Find the change of basis matrix $[I]_\alpha^\beta$.

Solution. (a) Expanding, $1+x$, $(1+x)^2 = 1+2x+x^2$ and $(1+x)^3 = 1+3x+3x^2+x^3$. Writing the coordinate vectors of the elements of $\alpha$ with respect to $\{1, x, x^2, x^3\}$ as columns gives

M = \begin{pmatrix} 1 & 1 & 1 & 1\\0 & 1 & 2 & 3\\0 & 0 & 1 & 3\\0 & 0 & 0 & 1\end{pmatrix}.

$M$ is upper triangular, so by property (vi) of the determinant from the lecture of September 23, $|M| = 1\not= 0$. By the Very Important Special Case of Column Vectors from the lecture of September 11, the columns of $M$ are linearly independent. Since coordinates respect linear combinations (formula (i) of the Three Important Formulas, lecture of September 18), the elements of $\alpha$ are linearly independent in $P(3)$. As $\dim(P(3)) = 4$ (Example (iii) in the lecture of September 9), the theorem in the lecture of September 11 shows that the four linearly independent elements of $\alpha$ form a basis.

(b) We seek $a, b, c, d$ with $p(x) = a+b(1+x)+c(1+x)^2+d(1+x)^3$. Comparing coefficients:

x^3:\ d = 1;\quad x^2:\ c+3d = 3;\quad x:\ b+2c+3d = -1;\quad 1:\ a+b+c+d = 2.

Back substitution gives $d = 1$, $c = 0$, $b = -4$, $a = 5$. Thus

[p(x)]_\alpha = \begin{pmatrix} 5\\-4\\0\\1\end{pmatrix}.

Check: $5-4(1+x)+(1+x)^3 = 5-4-4x+1+3x+3x^2+x^3 = 2-x+3x^2+x^3$. (Alternatively, substitute $x = y-1$ into $p$ to get $y^3-4y+5$.)

(c) By the definition of $[T]_\alpha^\beta$ given in the lecture of September 16 (applied to $T = I$, as in the lecture of September 18), the $j$th column of $[I]_\alpha^\beta$ is the coordinate vector with respect to $\beta$ of the $j$th element of $\alpha$, so $[I]_\alpha^\beta = M$ from part (a):

[I]_\alpha^\beta = \begin{pmatrix} 1 & 1 & 1 & 1\\0 & 1 & 2 & 3\\0 & 0 & 1 & 3\\0 & 0 & 0 & 1\end{pmatrix}.

As a check, using formula (ii) of the Three Important Formulas from the lecture of September 18,

[I]_\alpha^\beta\cdot[p]_\alpha = \begin{pmatrix} 5-4+0+1\\-4+0+3\\0+3\\1\end{pmatrix} = \begin{pmatrix} 2\\-1\\3\\1\end{pmatrix} = [p]_\beta.

Problem 7

Define $T: P(2)\to\mathbb{R}^3$ by

T(p(x)) = \begin{pmatrix} p(0)\\p(1)\\p'(1)\end{pmatrix}.

(a) Verify that $T$ is a linear transformation.
(b) Let $\alpha = \{1, x, x^2\}$ and $\beta = \left\{\begin{pmatrix} 1\\0\\0\end{pmatrix}, \begin{pmatrix} 1\\1\\0\end{pmatrix}, \begin{pmatrix} 1\\1\\1\end{pmatrix}\right\}$. Find $[T]_\alpha^\beta$.
(c) For $p(x) = 1-2x+x^2$, calculate $[T(p(x))]_\beta$ in two ways: directly, and by using the formula $[T(v)]_\beta = [T]_\alpha^\beta\cdot[v]_\alpha$.
(d) Use a determinant to decide whether $T$ is one-to-one and onto.

Solution. (a) We verify the two conditions in the definition of linear transformation given in the lecture of September 11. For $p, q\in P(2)$ and $\lambda\in\mathbb{R}$, using $(p+q)' = p'+q'$ and $(\lambda p)' = \lambda p'$,

\begin{aligned} T(p+q) &= \begin{pmatrix} p(0)+q(0)\\p(1)+q(1)\\p'(1)+q'(1)\end{pmatrix} = T(p)+T(q),\\[6pt] T(\lambda p) &= \begin{pmatrix} \lambda p(0)\\\lambda p(1)\\\lambda p'(1)\end{pmatrix} = \lambda T(p). \end{aligned}

(b) We have $T(1) = \begin{pmatrix} 1\\1\\0\end{pmatrix}$, $T(x) = \begin{pmatrix} 0\\1\\1\end{pmatrix}$, $T(x^2) = \begin{pmatrix} 0\\1\\2\end{pmatrix}$. By the definition of $[T]_\alpha^\beta$ given in the lecture of September 16, we must write these in terms of $\beta$. Writing $\beta = \{u_1, u_2, u_3\}$, a vector $\begin{pmatrix} a\\b\\c\end{pmatrix} = c_1u_1+c_2u_2+c_3u_3$ exactly when $c_3 = c$, $c_2 = b-c$, $c_1 = a-b$. Hence

T(1) = 0u_1+1u_2+0u_3,\quad T(x) = -u_1+0u_2+u_3,\quad T(x^2) = -u_1-u_2+2u_3,

and

[T]_\alpha^\beta = \begin{pmatrix} 0 & -1 & -1\\1 & 0 & -1\\0 & 1 & 2\end{pmatrix}.

(c) Directly: $p(x) = (x-1)^2$, so $p(0) = 1$, $p(1) = 0$, $p'(1) = 0$, and $T(p) = \begin{pmatrix} 1\\0\\0\end{pmatrix} = u_1$. Thus $[T(p)]_\beta = \begin{pmatrix} 1\\0\\0\end{pmatrix}$. Using formula (ii) of the Three Important Formulas from the lecture of September 18: $[p]_\alpha = \begin{pmatrix} 1\\-2\\1\end{pmatrix}$ and

[T]_\alpha^\beta\cdot[p]_\alpha = \begin{pmatrix} 0 & -1 & -1\\1 & 0 & -1\\0 & 1 & 2\end{pmatrix}\begin{pmatrix} 1\\-2\\1\end{pmatrix} = \begin{pmatrix} 0+2-1\\1+0-1\\0-2+2\end{pmatrix} = \begin{pmatrix} 1\\0\\0\end{pmatrix},

in agreement.

(d) Expanding along the first column, $\big|[T]_\alpha^\beta\big| = -1\cdot\begin{vmatrix} -1 & -1\\1 & 2\end{vmatrix} = -1\cdot(-2+1) = 1\not= 0$, so $[T]_\alpha^\beta$ is invertible by the theorem in the lecture of September 25. If $T(p) = \vec{0}$, then by formula (ii) of the Three Important Formulas (lecture of September 18), $[T]_\alpha^\beta[p]_\alpha = \vec{0}$, so $[p]_\alpha = \vec{0}$ and $p = 0$; thus $T$ is one-to-one (if $T(p) = T(q)$, apply this to $p-q$). Given $w\in\mathbb{R}^3$, set $X = ([T]_\alpha^\beta)^{-1}[w]_\beta$ and let $p$ be the polynomial with $[p]_\alpha = X$; then $[T(p)]_\beta = [w]_\beta$, so $T(p) = w$. Thus $T$ is onto.

Problem 8

Let $T: \mathbb{R}^2\to\mathbb{R}^3$ be the linear transformation satisfying

T\begin{pmatrix} 1\\1\end{pmatrix} = \begin{pmatrix} 1\\0\\2\end{pmatrix} \quad\text{and}\quad T\begin{pmatrix} 1\\-1\end{pmatrix} = \begin{pmatrix} 3\\2\\0\end{pmatrix},

and let $S: \mathbb{R}^3\to\mathbb{R}^2$ be given by $S\begin{pmatrix} x\\y\\z\end{pmatrix} = \begin{pmatrix} x-z\\y+2z\end{pmatrix}$.

(a) Find a formula for $T\begin{pmatrix} x\\y\end{pmatrix}$, and use it to calculate $T\begin{pmatrix} 5\\1\end{pmatrix}$.
(b) Find $[T]_{\epsilon_2}^{\epsilon_3}$, where $\epsilon_2$ and $\epsilon_3$ denote the standard bases of $\mathbb{R}^2$ and $\mathbb{R}^3$.
(c) Let $\alpha = \left\{\begin{pmatrix} 1\\1\end{pmatrix}, \begin{pmatrix} 1\\-1\end{pmatrix}\right\}$. Write down $[T]_\alpha^{\epsilon_3}$ and $[S]_{\epsilon_3}^{\epsilon_2}$, and use them to calculate $[ST]_\alpha^{\epsilon_2}$.
(d) Check your answer to (c) by calculating $ST\begin{pmatrix} 1\\1\end{pmatrix}$ and $ST\begin{pmatrix} 1\\-1\end{pmatrix}$ directly.

Solution. (a) Solving $\begin{pmatrix} x\\y\end{pmatrix} = a\begin{pmatrix} 1\\1\end{pmatrix}+b\begin{pmatrix} 1\\-1\end{pmatrix}$ gives $a = \frac{x+y}{2}$, $b = \frac{x-y}{2}$. Since a linear transformation is determined by its effect on a basis (lecture of September 16),

T\begin{pmatrix} x\\y\end{pmatrix} = \frac{x+y}{2}\begin{pmatrix} 1\\0\\2\end{pmatrix}+\frac{x-y}{2}\begin{pmatrix} 3\\2\\0\end{pmatrix} = \begin{pmatrix} 2x-y\\x-y\\x+y\end{pmatrix}.

In particular, $T\begin{pmatrix} 5\\1\end{pmatrix} = \begin{pmatrix} 9\\4\\6\end{pmatrix}$. (Check: this formula gives the stated values of $T$ at $\begin{pmatrix} 1\\1\end{pmatrix}$ and $\begin{pmatrix} 1\\-1\end{pmatrix}$.)

(b) By the definition of $[T]_\alpha^\beta$ given in the lecture of September 16, the columns are $T(e_1)$ and $T(e_2)$:

[T]_{\epsilon_2}^{\epsilon_3} = \begin{pmatrix} 2 & -1\\1 & -1\\1 & 1\end{pmatrix}.

(c) The columns of $[T]_\alpha^{\epsilon_3}$ are the given values of $T$ on the elements of $\alpha$, and the columns of $[S]_{\epsilon_3}^{\epsilon_2}$ are $S(e_1), S(e_2), S(e_3)$:

[T]_\alpha^{\epsilon_3} = \begin{pmatrix} 1 & 3\\0 & 2\\2 & 0\end{pmatrix}, \qquad [S]_{\epsilon_3}^{\epsilon_2} = \begin{pmatrix} 1 & 0 & -1\\0 & 1 & 2\end{pmatrix}.

By formula (iii) of the Three Important Formulas from the lecture of September 18,

[ST]_\alpha^{\epsilon_2} = [S]_{\epsilon_3}^{\epsilon_2}\cdot[T]_\alpha^{\epsilon_3} = \begin{pmatrix} 1 & 0 & -1\\0 & 1 & 2\end{pmatrix}\begin{pmatrix} 1 & 3\\0 & 2\\2 & 0\end{pmatrix} = \begin{pmatrix} -1 & 3\\4 & 2\end{pmatrix}.

(d) $ST\begin{pmatrix} 1\\1\end{pmatrix} = S\begin{pmatrix} 1\\0\\2\end{pmatrix} = \begin{pmatrix} 1-2\\0+4\end{pmatrix} = \begin{pmatrix} -1\\4\end{pmatrix}$ and $ST\begin{pmatrix} 1\\-1\end{pmatrix} = S\begin{pmatrix} 3\\2\\0\end{pmatrix} = \begin{pmatrix} 3\\2\end{pmatrix}$. These are the columns found in (c).

Problem 9

Let $T: \mathbb{R}^2\to\mathbb{R}^2$ be given by $T\begin{pmatrix} x\\y\end{pmatrix} = \begin{pmatrix} 4x-2y\\x+y\end{pmatrix}$. Let $\alpha = \{e_1, e_2\}$ be the standard basis and $\beta = \{f_1, f_2\}$, where $f_1 = \begin{pmatrix} 2\\1\end{pmatrix}$ and $f_2 = \begin{pmatrix} 1\\1\end{pmatrix}$.

(a) Find $[T]_\alpha^\alpha$.
(b) Find the change of basis matrices $[I]_\beta^\alpha$ and $[I]_\alpha^\beta$, and verify that their product is $I_2$.
(c) Use the change of basis formula to calculate $[T]_\beta^\beta$.
(d) Check your answer to (c) by calculating $T(f_1)$ and $T(f_2)$ directly and expressing them in terms of $f_1, f_2$.

Solution. (a) $T(e_1) = \begin{pmatrix} 4\\1\end{pmatrix}$ and $T(e_2) = \begin{pmatrix} -2\\1\end{pmatrix}$, so $A := [T]_\alpha^\alpha = \begin{pmatrix} 4 & -2\\1 & 1\end{pmatrix}$.

(b) The columns of $[I]_\beta^\alpha$ are $f_1, f_2$ written in the standard basis, so $P := [I]_\beta^\alpha = \begin{pmatrix} 2 & 1\\1 & 1\end{pmatrix}$. By the Observations at the end of the lecture of September 18, $[I]_\alpha^\beta$ is its inverse; since $|P| = 1$, using $P^{-1} = \frac{1}{|P|}P'$ from the lecture of September 25,

[I]_\alpha^\beta = P^{-1} = \begin{pmatrix} 1 & -1\\-1 & 2\end{pmatrix}.

(Directly: $e_1 = f_1-f_2$ and $e_2 = -f_1+2f_2$.) Check: $\begin{pmatrix} 2 & 1\\1 & 1\end{pmatrix}\begin{pmatrix} 1 & -1\\-1 & 2\end{pmatrix} = \begin{pmatrix} 1 & 0\\0 & 1\end{pmatrix}$, and similarly in the other order.

(c) By the Change of Basis Formula from the lecture of September 21,

\begin{aligned} [T]_\beta^\beta &= [I]_\alpha^\beta\,[T]_\alpha^\alpha\,[I]_\beta^\alpha = P^{-1}AP = \begin{pmatrix} 1 & -1\\-1 & 2\end{pmatrix}\begin{pmatrix} 4 & -2\\1 & 1\end{pmatrix}\begin{pmatrix} 2 & 1\\1 & 1\end{pmatrix}\\[6pt] &= \begin{pmatrix} 1 & -1\\-1 & 2\end{pmatrix}\begin{pmatrix} 6 & 2\\3 & 2\end{pmatrix} = \begin{pmatrix} 3 & 0\\0 & 2\end{pmatrix}. \end{aligned}

(d) $T(f_1) = \begin{pmatrix} 8-2\\2+1\end{pmatrix} = \begin{pmatrix} 6\\3\end{pmatrix} = 3f_1+0f_2$ and $T(f_2) = \begin{pmatrix} 4-2\\1+1\end{pmatrix} = \begin{pmatrix} 2\\2\end{pmatrix} = 0f_1+2f_2$. So $[T]_\beta^\beta = \begin{pmatrix} 3 & 0\\0 & 2\end{pmatrix}$, as in (c).

Problem 10

Determinants.

(a) Use elementary row operations to calculate $\begin{vmatrix} 1 & 2 & 0 & 3\\2 & 5 & 1 & 4\\-1 & 0 & 3 & 2\\3 & 6 & 1 & 10\end{vmatrix}$.
(b) Let $B = \begin{pmatrix} 1 & 0 & 2\\2 & 1 & 0\\0 & 3 & 1\end{pmatrix}$. Calculate $|B|$, the classical adjoint $B'$ of $B$, and use these to find $B^{-1}$.
(c) Use Cramer's Rule to find $x_2$ in the solution to the system $\begin{array}{rcrcrcr} x_1 & & & + & 2x_3 & = & 1\\ 2x_1 & + & x_2 & & & = & 4\\ & & 3x_2 & + & x_3 & = & 2. \end{array}$
(d) Suppose $A$ is a $3\times 3$ matrix with $|A| = 3$. Calculate $|2A|$, $|A^{-1}|$, $|A^4|$ and $|A'|$, where $A'$ denotes the classical adjoint of $A$.

Solution. (a) By property (v) of the determinant from the lecture of September 23, adding a multiple of one row to another row does not change the determinant. Using $R_2-2R_1$, $R_3+R_1$, $R_4-3R_1$, and then $R_3-2R_2$, $R_4-R_3$:

\begin{aligned} \begin{vmatrix} 1 & 2 & 0 & 3\\2 & 5 & 1 & 4\\-1 & 0 & 3 & 2\\3 & 6 & 1 & 10\end{vmatrix} &= \begin{vmatrix} 1 & 2 & 0 & 3\\0 & 1 & 1 & -2\\0 & 2 & 3 & 5\\0 & 0 & 1 & 1\end{vmatrix} = \begin{vmatrix} 1 & 2 & 0 & 3\\0 & 1 & 1 & -2\\0 & 0 & 1 & 9\\0 & 0 & 1 & 1\end{vmatrix}\\[6pt] &= \begin{vmatrix} 1 & 2 & 0 & 3\\0 & 1 & 1 & -2\\0 & 0 & 1 & 9\\0 & 0 & 0 & -8\end{vmatrix} = 1\cdot 1\cdot 1\cdot(-8) = -8, \end{aligned}

the last step using property (vi) from the lecture of September 23, since the last matrix is upper triangular.

(b) Expanding along the first row, as in the definition of the determinant from the lecture of September 23, $|B| = 1(1\cdot 1-0\cdot 3)-0+2(2\cdot 3-1\cdot 0) = 1+12 = 13$. The cofactors $(-1)^{i+j}|B_{ij}|$ are

\begin{align*} &C_{11} = 1, && C_{12} = -2, && C_{13} = 6,\\ &C_{21} = -(0-6) = 6, && C_{22} = 1, && C_{23} = -(3-0) = -3,\\ &C_{31} = 0-2 = -2, && C_{32} = -(0-4) = 4, && C_{33} = 1. \end{align*}

By the definition in the lecture of September 25, the classical adjoint is the transpose of the matrix of cofactors, and, as shown in that lecture, $B^{-1} = \frac{1}{|B|}B'$:

B' = \begin{pmatrix} 1 & 6 & -2\\-2 & 1 & 4\\6 & -3 & 1\end{pmatrix}, \qquad B^{-1} = \frac{1}{13}\begin{pmatrix} 1 & 6 & -2\\-2 & 1 & 4\\6 & -3 & 1\end{pmatrix}.

(Check: the first row of $B$ times the columns of $B'$ gives $(1+12,\ 6-6,\ -2+2) = (13, 0, 0)$, etc.)

(c) The coefficient matrix of the system is $B$ from part (b), and $b = \begin{pmatrix} 1\\4\\2\end{pmatrix}$. Replacing the second column of $B$ by $b$,

|B_2| = \begin{vmatrix} 1 & 1 & 2\\2 & 4 & 0\\0 & 2 & 1\end{vmatrix} = 1(4-0)-1(2-0)+2(4-0) = 10,

so by Cramer's Rule, from the lecture of September 25, $x_2 = \frac{|B_2|}{|B|} = \frac{10}{13}$. (In fact, $X = B^{-1}b = \frac{1}{13}\begin{pmatrix} 21\\10\\-4\end{pmatrix}$.)

(d) By property (i) of the determinant from the lecture of September 23, multiplying each of the three rows of $A$ by 2 multiplies the determinant by 2 three times, so $|2A| = 2^3|A| = 24$. From $AA^{-1} = I_3$ and $|AB| = |A||B|$ (lecture of September 25), $|A||A^{-1}| = 1$, so $|A^{-1}| = \frac{1}{3}$. Similarly $|A^4| = |A|^4 = 81$. Finally, $AA' = |A|\cdot I_3$ (lecture of September 25) gives $|A|\cdot|A'| = \big||A|\cdot I_3\big| = |A|^3$, so $|A'| = |A|^2 = 9$.