# 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*)

Contents

- Problem 389
- Solution.
- Solution 1 of (a) A $2 \times 2$ matrix $A$ satisfies $\tr(A^2)=5$ and $\tr(A)=3$. Find $\det(A)$.
- Solution 2 of (a)
- Solution 1 of (b) A $2 \times 2$ matrix has two parallel columns and $\tr(A)=5$. Find $\tr(A^2)$.
- Solution 2 of (b)
- Solution of (c) A $2\times 2$ matrix $A$ has $\det(A)=5$ and positive integer eigenvalues. What is the trace of $A$?

## 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

\[\tr(A)=1+5=6.\]

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