Exercises — Lesson 14: Determinants and Trace
Exercise 14.1. [Hand] Compute by row reduction, tracking swaps and column operations’ effects: \(\det\begin{pmatrix}1&2&0\\2&4&1\\1&0&3\end{pmatrix}\). Cross-check with the \(3\times3\) formula.
Solution.
\[ \det\begin{pmatrix} 1 & 2 & 0 \\ 2 & 4 & 1 \\ 1 & 0 & 3 \end{pmatrix} \;\longrightarrow\; \begin{pmatrix} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & -2 & 3 \end{pmatrix} \;\longrightarrow\; \begin{pmatrix} 1 & 2 & 0 \\ 0 & -2 & 3 \\ 0 & 0 & 1 \end{pmatrix} \]
\[ \begin{aligned} &= -\big[(1)(-2)(1)\big] \\[6pt] &= \boxed{\;2\;} \end{aligned} \]
Cross-check with the \(3\times3\) formula:
\[ 1\begin{pmatrix} 4 & 1 \\ 0 & 3 \end{pmatrix} - 2\begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} + 0\begin{pmatrix} 2 & 4 \\ 1 & 0 \end{pmatrix} = 1(12) - 2(5) + 0 = 2 \]
Exercise 14.2. [Hand] Using trace and det only, find the eigenvalues of \(\left(\begin{smallmatrix}5&4\\1&2\end{smallmatrix}\right)\) and \(\left(\begin{smallmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{smallmatrix}\right)\) (the latter over \(\mathbb{C}\); interpret).
Solution.
\[ \begin{aligned} \operatorname{tr}\begin{pmatrix} 5 & 4 \\ 1 & 2 \end{pmatrix} &= 7 \\[6pt] \det\begin{pmatrix} 5 & 4 \\ 1 & 2 \end{pmatrix} &= 6 \end{aligned} \qquad\qquad \begin{aligned} \operatorname{tr}\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} &= 2\cos\theta \\[6pt] \det\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} &= \cos^2\theta + \sin^2\theta = 1 \end{aligned} \]
\[ \frac{\operatorname{tr}A \pm \sqrt{(\operatorname{tr}A)^2 - 4\det A}}{2} \]
For \(\begin{pmatrix} 5 & 4 \\ 1 & 2 \end{pmatrix}\):
\[ \lambda = \frac{7 \pm \sqrt{49 - 24}}{2} = \frac{7 \pm \sqrt{25}}{2} = \frac{7 \pm 5}{2} \qquad \boxed{\; \begin{aligned} \lambda_1 &= 6 \\[4pt] \lambda_2 &= 1 \end{aligned} \;} \]
For the rotation matrix:
\[ \begin{aligned} \frac{2\cos\theta \pm \sqrt{4\cos^2\theta - 4}}{2} &\;\Longrightarrow\; \cos\theta \pm \sqrt{\cos^2\theta - 1} \\[6pt] &\;\Longrightarrow\; \cos\theta \pm \sqrt{-1\left(1 - \cos^2\theta\right)} \\[6pt] &\;\Longrightarrow\; \cos\theta \pm \sin\theta\sqrt{-1} \\[6pt] &\;\Longrightarrow\; \cos\theta \pm i\sin\theta \end{aligned} \qquad \boxed{\; \begin{aligned} \lambda_1 &= e^{i\theta} \\[4pt] \lambda_2 &= e^{-i\theta} \end{aligned} \;} \]
Exercise 14.3. [Proof, \(\star\)] From the three defining properties alone (no permutation formula), derive \(\det\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right) = ad - bc\): expand \(\det\big((a,c),(b,d)\big)\) by multilinearity into four determinants of standard-basis columns, and evaluate each using alternation and normalization.
Proof. Let \(e_1, e_2\) be standard basis vectors.
\[ \begin{aligned} \det\big((a,c),\,(b,d)\big) &= \det\big(ae_1 + ce_2,\; be_1 + de_2\big) \\[6pt] &= a\det\big(e_1,\; be_1 + de_2\big) + c\det\big(e_2,\; be_1 + de_2\big) \\[6pt] &= ab\det(e_1, e_1) + ad\det(e_1, e_2) + cb\det(e_2, e_1) + cd\det(e_2, e_2) \\[6pt] &= ad\det(e_1, e_2) + bc\det(e_2, e_1) \\[6pt] &= ad\det(e_1, e_2) - bc\det(e_1, e_2) \\[6pt] &= ad - bc \end{aligned} \]
\(\square\)
Exercise 14.4. [Proof] Prove that similar matrices (\(B = S^{-1}AS\)) have the same determinant, trace, and eigenvalues — so all three are properties of the underlying linear map, independent of basis.
Proof.
\[ \det B = \det(S^{-1}AS) = \det(S^{-1})\det(A)\det(S) = \det(S^{-1})\det(S)\det(A) = \det(S^{-1}S)\det A = \det(I)\det(A) = \det(A) \]
\[ \operatorname{tr}B = \operatorname{tr}(S^{-1}AS) = \operatorname{tr}\big(S^{-1}(AS)\big) = \operatorname{tr}\big((AS)S^{-1}\big) = \operatorname{tr}\big(A(SS^{-1})\big) = \operatorname{tr}A \]
\[ \begin{aligned} \det(B - \lambda I) &= \det(S^{-1}AS - \lambda I) \\[6pt] &= \det(S^{-1}AS - \lambda S^{-1}S) \\[6pt] &= \det\big(S^{-1}(A - \lambda I)S\big) && \left[-\lambda S^{-1}S = S^{-1}(-\lambda I)S\right] \\[6pt] &= \det(S^{-1})\det(A - \lambda I)\det(S) \\[6pt] &= \det(S^{-1})\det(S)\det(A - \lambda I) \\[6pt] &= \det(S^{-1}S)\det(A - \lambda I) \\[6pt] &= \det(A - \lambda I) \end{aligned} \]
So \(A\) and \(B = (S^{-1}AS)\) have same determinant, trace, and eigenvalues. \(\square\)