Fall 2026: Math 590 Exam 1 Solutions

Throughout, unless \(V\) is identified explicitly, \(V\) will denote a vector space over \(F\), where \(F\) denotes either \(\mathbb{R}\) or \(\mathbb{C}\).

True-False and Short Answer (40 points)

Problem 1

True or False: If \(V\) is a 10 dimensional vector space, then \(V\) can be spanned by 15 vectors.

Answer. True.

Problem 2

True or False: There exists a system of \(m\) linear equations in \(n\) unknowns with exactly 17 solutions.

Answer. False.

Problem 3

Define what it means for the vectors \(v_1, \ldots, v_n\in V\) to be linearly independent. Do not just say the vectors are not linearly dependent.

Solution. The vectors \(v_1, \ldots, v_n\) are linearly independent if whenever \(a_1, \ldots, a_n\in F\) satisfy

\[a_1v_1+a_2v_2+\cdots+a_nv_n = 0,\]

then \(a_1 = a_2 = \cdots = a_n = 0\). Equivalently, no \(v_j\) belongs to \(\mathrm{Span}\{v_1, \ldots, v_{j-1}, v_{j+1}, \ldots, v_n\}\).

Problem 4

State the Exchange Theorem.

Solution. Suppose \(v_1, \ldots, v_t\) span the subspace \(W\) and \(w_1, \ldots, w_s\in W\) are linearly independent. Then \(s\leq t\), and after renumbering the \(v_i\) if necessary, the vectors

\[w_1, \ldots, w_s, v_{s+1}, \ldots, v_t\]

span \(W\).

Problem 5

Let \(A\) be an \(n\times n\) matrix. State three equivalent conditions for \(|A|\not= 0\).

Solution. Possible answers include the following which are equivalent to \(|A|\neq 0\):

  1. (i) \(A\) is invertible.
  2. (ii) The rows (or columns) of \(A\) form a basis for \(F^n\).
  3. (iii) For every \(b\in F^n\), the system \(AX = b\) has a unique solution.
  4. (iv) The columns (equivalently, the rows) of \(A\) are linearly independent.
  5. (v) \(A\) is row equivalent to the identity matrix \(I_n\).

Problem 6

Write \(\mathrm{det}\begin{pmatrix} 1 & 0 & 3 & -4\\2 & 1 & 0 & -6\\0 & 0 & 2 & 5\\1 & 1 & 2 & 9\end{pmatrix}\) by expanding along the 3rd column. Do not calculate the resulting \(3\times 3\) determinants.

Solution. The entries of the third column are \(a_{13} = 3\), \(a_{23} = 0\), \(a_{33} = 2\), \(a_{43} = 2\), with signs \((-1)^{i+3}\), namely \(+, -, +, -\). Thus

\[\begin{aligned} \mathrm{det}\begin{pmatrix} 1 & 0 & 3 & -4\\2 & 1 & 0 & -6\\0 & 0 & 2 & 5\\1 & 1 & 2 & 9\end{pmatrix} &= 3\,\mathrm{det}\begin{pmatrix} 2 & 1 & -6\\0 & 0 & 5\\1 & 1 & 9\end{pmatrix} - 0\,\mathrm{det}\begin{pmatrix} 1 & 0 & -4\\0 & 0 & 5\\1 & 1 & 9\end{pmatrix}\\[6pt] &\quad + 2\,\mathrm{det}\begin{pmatrix} 1 & 0 & -4\\2 & 1 & -6\\1 & 1 & 9\end{pmatrix} - 2\,\mathrm{det}\begin{pmatrix} 1 & 0 & -4\\2 & 1 & -6\\0 & 0 & 5\end{pmatrix}. \end{aligned}\]

Problem 7

True or False: \(\mathrm{det}\begin{pmatrix} a & b & c\\d & e & f\\a & b & c\end{pmatrix} = -\mathrm{det}\begin{pmatrix} a & b & c\\d & e & f\\a & b & c\end{pmatrix}\).

Answer. True.

Problem 8

Let \(V, W, U\) be vector spaces over \(F\) with bases \(\alpha, \beta, \gamma\) and suppose \(T: V\to W\) and \(S: W\to U\) are linear transformations. Place the given bases in the appropriate places to make the formula \([ST] = [S]\cdot [T]\) valid.

Solution.

\[[ST]_{\alpha}^{\gamma} = [S]_{\beta}^{\gamma}\cdot [T]_{\alpha}^{\beta}.\]

Problem 9

Let \(T: V\to V\) be a linear transformation and suppose \(\alpha, \beta\) are bases for \(V\). Write the change of basis formula relating the matrix of \(T\) with respect to \(\alpha\) to the matrix of \(T\) with respect to \(\beta\).

Solution. Let \(I: V\to V\) be the identity map. Then

\[[T]_{\beta}^{\beta} = [I]_{\alpha}^{\beta}\,[T]_{\alpha}^{\alpha}\,[I]_{\beta}^{\alpha}.\]

Problem 10

For the matrix \(A = \begin{pmatrix} 2 & 4 & -9\\0 & 2 & 5\\1 & 5 & 2\end{pmatrix}\), find the \((2,3)\) entry of \(A'\), the adjugate or classical adjoint of \(A\).

Solution. The adjugate is the transpose of the cofactor matrix \(C\) whose \((i,j)\) entry is \((-1)^{i+j}|A_{ij}|\), so the \((2,3)\) entry of \(A'\) is the \((3,2)\) entry of \(C\). Deleting row 3 and column 2 of \(A\) gives

\[(A')_{23} = C_{32} = (-1)^{3+2}\,\mathrm{det}\begin{pmatrix} 2 & -9\\0 & 5\end{pmatrix} = -(10-0) = -10.\]

Calculation Problems (60 points)

For these problems, we let \(V\) denote the vector space of real \(2\times 2\) matrices and recall that if \(A = \begin{pmatrix} a & b\\c & d\end{pmatrix}\), then \(\mathrm{tr}(A)\), the trace of \(A\), is \(a+d\).

Problem 1

Let \(W = \{A\in V\ |\ \mathrm{tr}(A) = 0\}\). Verify that \(W\) is a subspace of \(V\).

Solution. We check the two subspace conditions. Let \(A = \begin{pmatrix} a_1 & b_1\\c_1 & d_1\end{pmatrix}\) and \(B = \begin{pmatrix} a_2 & b_2\\c_2 & d_2\end{pmatrix}\) be in \(W\), so \(a_1+d_1 = 0\) and \(a_2+d_2 = 0\), and let \(\lambda\in\mathbb{R}\).

  1. (i) \(A+B = \begin{pmatrix} a_1+a_2 & b_1+b_2\\c_1+c_2 & d_1+d_2\end{pmatrix}\), so \(\mathrm{tr}(A+B) = (a_1+a_2)+(d_1+d_2) = (a_1+d_1)+(a_2+d_2) = 0+0 = 0\). Hence \(A+B\in W\), so \(W\) is closed under addition.
  2. (ii) \(\lambda A = \begin{pmatrix} \lambda a_1 & \lambda b_1\\\lambda c_1 & \lambda d_1\end{pmatrix}\), so \(\mathrm{tr}(\lambda A) = \lambda a_1+\lambda d_1 = \lambda(a_1+d_1) = \lambda\cdot 0 = 0\). Hence \(\lambda A\in W\), so \(W\) is closed under scalar multiplication.

Therefore \(W\) is a subspace of \(V\).

Problem 2

Find a basis \(\alpha\) for \(W\). Be sure to justify your answer.

Solution. If \(A = \begin{pmatrix} a & b\\c & d\end{pmatrix}\in W\), then \(d = -a\), so

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

Let

\[v_1 = \begin{pmatrix} 1 & 0\\0 & -1\end{pmatrix},\quad v_2 = \begin{pmatrix} 0 & 1\\0 & 0\end{pmatrix},\quad v_3 = \begin{pmatrix} 0 & 0\\1 & 0\end{pmatrix},\]

and \(\alpha = \{v_1, v_2, v_3\}\). Each \(v_i\) has trace \(0\), so \(\alpha\subseteq W\).

Spanning: The displayed equation shows every \(A\in W\) is a linear combination of \(v_1, v_2, v_3\), so \(\alpha\) spans \(W\).

Linear independence: If \(av_1+bv_2+cv_3 = 0\), then

\[\begin{pmatrix} a & b\\c & -a\end{pmatrix} = \begin{pmatrix} 0 & 0\\0 & 0\end{pmatrix},\]

so \(a = b = c = 0\).

Hence \(\alpha\) is a basis for \(W\), and \(\dim W = 3\).

Problem 3

For \(A = \begin{pmatrix} a & b\\c & d\end{pmatrix}\) in \(W\), let \(T: W\to W\) be the linear transformation defined by \(T(A) = A'\), where \(A'\) is the adjugate or classical adjoint of \(A\). For \(\alpha\) in Problem 2, find \([T]_{\alpha}^{\alpha}\).

Solution. For a \(2\times 2\) matrix, \(\begin{pmatrix} a & b\\c & d\end{pmatrix}' = \begin{pmatrix} d & -b\\-c & a\end{pmatrix}\). Applying this to the basis vectors:

\[\begin{align*} T(v_1) &= \begin{pmatrix} -1 & 0\\0 & 1\end{pmatrix} = -1\cdot v_1+0\cdot v_2+0\cdot v_3,\\ T(v_2) &= \begin{pmatrix} 0 & -1\\0 & 0\end{pmatrix} = 0\cdot v_1-1\cdot v_2+0\cdot v_3,\\ T(v_3) &= \begin{pmatrix} 0 & 0\\-1 & 0\end{pmatrix} = 0\cdot v_1+0\cdot v_2-1\cdot v_3. \end{align*}\]

Thus:

\[[T]_{\alpha}^{\alpha} = \begin{pmatrix} -1 & 0 & 0\\0 & -1 & 0\\0 & 0 & -1\end{pmatrix} = -I_3.\]

Problem 4

For \(A\in W\), let \(S: W\to V\) be defined by \(S(A) = A\). Let \(\beta\subseteq V\) be the standard basis. Find \([ST]_{\alpha}^{\beta}\) in two ways: directly and by using the product formula.

Solution. Here \(S(A) = A\) is the inclusion of \(W\) into \(V\), and \(\beta = \{E_1, E_2, E_3, E_4\}\), where

\[E_1 = \begin{pmatrix} 1 & 0\\0 & 0\end{pmatrix},\quad E_2 = \begin{pmatrix} 0 & 1\\0 & 0\end{pmatrix},\quad E_3 = \begin{pmatrix} 0 & 0\\1 & 0\end{pmatrix},\quad E_4 = \begin{pmatrix} 0 & 0\\0 & 1\end{pmatrix}.\]

Directly. By Problem 3, \(ST(A) = S(-A) = -A\) for \(A\in W\). Thus

\[\begin{align*} ST(v_1) &= \begin{pmatrix} -1 & 0\\0 & 1\end{pmatrix} = -E_1+0E_2+0E_3+E_4,\\ ST(v_2) &= \begin{pmatrix} 0 & -1\\0 & 0\end{pmatrix} = 0E_1-E_2+0E_3+0E_4,\\ ST(v_3) &= \begin{pmatrix} 0 & 0\\-1 & 0\end{pmatrix} = 0E_1+0E_2-E_3+0E_4, \end{align*}\]

so

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

Using the product formula. First compute \([S]_{\alpha}^{\beta}\):

\[\begin{align*} S(v_1) &= \begin{pmatrix} 1 & 0\\0 & -1\end{pmatrix} = E_1+0E_2+0E_3-E_4,\\ S(v_2) &= \begin{pmatrix} 0 & 1\\0 & 0\end{pmatrix} = 0E_1+E_2+0E_3+0E_4,\\ S(v_3) &= \begin{pmatrix} 0 & 0\\1 & 0\end{pmatrix} = 0E_1+0E_2+E_3+0E_4, \end{align*}\]

so

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

Then

\[[ST]_{\alpha}^{\beta} = [S]_{\alpha}^{\beta}\,[T]_{\alpha}^{\alpha} = \begin{pmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\\-1 & 0 & 0\end{pmatrix}\begin{pmatrix} -1 & 0 & 0\\0 & -1 & 0\\0 & 0 & -1\end{pmatrix} = \begin{pmatrix} -1 & 0 & 0\\0 & -1 & 0\\0 & 0 & -1\\1 & 0 & 0\end{pmatrix},\]

which agrees with the direct computation.

Bonus Problem (10 points)

Let \(T: V\to W\) be a linear transformation and suppose \(\alpha, \alpha'\) are bases for \(V\) and \(\beta, \beta'\) are bases for \(W\). Find a change of basis formula relating the matrix of \(T\) with respect to \(\alpha, \beta\) to the matrix of \(T\) with respect to \(\alpha', \beta'\).

Solution. Let \(I_V: V\to V\) and \(I_W: W\to W\) be the identity maps. Then \(T = I_W\circ T\circ I_V\). Viewing \(I_V\) as a map from \(V\) with basis \(\alpha'\) to \(V\) with basis \(\alpha\), \(T\) as a map from \(V\) with basis \(\alpha\) to \(W\) with basis \(\beta\), and \(I_W\) as a map from \(W\) with basis \(\beta\) to \(W\) with basis \(\beta'\), the product formula from Problem 8 (applied twice) gives

\[[T]_{\alpha'}^{\beta'} = [I_W T I_V]_{\alpha'}^{\beta'} = [I_W]_{\beta}^{\beta'}\,[T]_{\alpha}^{\beta}\,[I_V]_{\alpha'}^{\alpha}.\]