Trace, Determinant, and Eigenvalue (Harvard University Exam Problem)

Problem 389

(a) A $2 \times 2$ matrix $A$ satisfies $\tr(A^2)=5$ and $\tr(A)=3$.
Find $\det(A)$.

(b) A $2 \times 2$ matrix has two parallel columns and $\tr(A)=5$. Find $\tr(A^2)$.

(c) A $2\times 2$ matrix $A$ has $\det(A)=5$ and positive integer eigenvalues. What is the trace of $A$?

(Harvard University, Linear Algebra Exam Problem)

 
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Solution.

For (a) and (b), we give two solutions. The first one does not use the knowledge of eigenvalues, and the second one uses eigenvalues.

Solution 1 of (a) A $2 \times 2$ matrix $A$ satisfies $\tr(A^2)=5$ and $\tr(A)=3$. Find $\det(A)$.

Let
\[A=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}.\] Then we have
\begin{align*}
A^2=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}=\begin{bmatrix}
a^2+bc & ab+bd\\
ac+cd& bc+d^2
\end{bmatrix}.
\end{align*}
Since $\tr(A^2)=5$ and $\tr(A)=3$, we obtain
\begin{align*}
5&=\tr(A^2)=(a^2+bc)+(bc+d^2)=a^2+2bc+d^2 \text{ and }\\
3&=\tr(A)=a+d.
\end{align*}
We find the determinant $\det(A)=ad-bc$ as follows.
We have
\begin{align*}
\det(A)&=ad-bc=\frac{1}{2}\left(\, (a+d)^2-(a^2+2bc+d^2) \,\right)\\
&=\frac{1}{2}(3^2-5)=2.
\end{align*}
Thus, we obtain $\det(A)=2$.

Solution 2 of (a)

Let $\lambda_1$ and $\lambda_2$ be eigenvalues of $A$.
Then we have
\begin{align*}
3=\tr(A)=\lambda_1+\lambda_2 \text{ and }\\
5=\tr(A^2)=\lambda_1^2+\lambda_2^2.
\end{align*}
Here we used two facts.
The first one is that the trace of a matrix is the sum of all eigenvalues of the matrix.
The second one is that $\lambda^2$ is an eigenvalue of $A^2$ if $\lambda$ is an eigenvalue of $A$, and these are all the eigenvalues of $A^2$.

Since the determinant of $A$ is the product of eigenvalues of $A$, we have
\begin{align*}
\det(A)&=\lambda_1 \lambda_2\\
&=\frac{1}{2}\left(\, (\lambda_1+\lambda_2)^2-(\lambda_1^2+\lambda_2^2) \,\right)\\
&=\frac{1}{2}(3^2-5)\\
&=2.
\end{align*}
Hence we have $\det(A)=2$.

Solution 1 of (b) A $2 \times 2$ matrix has two parallel columns and $\tr(A)=5$. Find $\tr(A^2)$.

Since two columns are parallel, we can write $A$ as
\[A=\begin{bmatrix}
a & ra\\
c& rc
\end{bmatrix}.\] Then we have
\begin{align*}
5=\tr(A)=a+rc.
\end{align*}
We use the formula in Solution 1 of (a) for $\tr(A^2)$ with $b=ra$ and $d=rc$, and we compute
\begin{align*}
\tr(A^2)&=a^2+2(ra)+(rc)^2\\
&=(a+rc)^2\\
&=5^2=25.
\end{align*}
Thus, we find $\tr(A^2)=25$.

Solution 2 of (b)

Since two columns are parallel, the matrix $A$ is singular. Hence $A$ has an eigenvalue $0$.
Since the sum of all the eigenvalues is $\tr(A)=5$, we see that $0$ and $5$ are eigenvalues of $A$.
It follows that $0$ and $25$ are eigenvalues of $A^2$. Hence
\[\tr(A^2)=0+25=25.\]

Solution of (c) A $2\times 2$ matrix $A$ has $\det(A)=5$ and positive integer eigenvalues. What is the trace of $A$?

The product of eigenvalues of $A$ is the determinant $\det(A)=5$.
Since eigenvalues are positive integers, it follows that $1$ and $5$ are eigenvalues of $A$.
It follows that
\[\tr(A)=1+5=6.\] Click here if solved 99

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