(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)

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 71

The post Trace, Determinant, and Eigenvalue (Harvard University Exam Problem) first appeared on Problems in Mathematics.

Find the determinant of the following matrix
\[A=\begin{bmatrix}
6 & 2 & 2 & 2 &2 \\
2 & 6 & 2 & 2 & 2 \\
2 & 2 & 6 & 2 & 2 \\
2 & 2 & 2 & 6 & 2 \\
2 & 2 & 2 & 2 & 6
\end{bmatrix}.\]

(Harvard University, Linear Algebra Exam Problem)
 

Hint.

Computing the determinant directly by hand is tedious.
So use the fact that the determinant of a matrix $A$ is the product of all eigenvalues of $A$.

However, finding the eigenvalue of $A$ itself is as complicated as computing the determinant of $A$.
Instead, first determine the eigenvalues of $B=A-4I$.
Then use the fact that if $\lambda$ is an eigenvalue of $B$, then $\lambda+4$ is an eigenvalue of $A$.

For a proof of this fact, see the post “Eigenvalues and algebraic/geometric multiplicities of matrix $A+cI$“.

Solution.

Let $B=A-4I$, where $I$ is the $5 \times 5$ identity matrix.
Then every entry of $B$ is $2$.

By elementary row operations, we can reduced the matrix $B$ into
\[B=\begin{bmatrix}
2 & 2 & 2 & 2 &2 \\
2 & 2 & 2 & 2 & 2 \\
2 & 2 & 2 & 2 & 2 \\
2 & 2 & 2 & 2 & 2 \\
2 & 2 & 2 & 2 & 2
\end{bmatrix}\to
\begin{bmatrix}
1 & 1 & 1 & 1 &1 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}.\] Thus, the rank of $B$ is $1$, hence the nullity is $4$ by the rank-nullity theorem.
It follows that $0$ is an eigenvalue of $B$ and its geometric multiplicity is $4$.

Since all entries of $B$ are equal, we compute
\[B\begin{bmatrix}
1 \\
1 \\
1 \\
1 \\
1
\end{bmatrix}=10\begin{bmatrix}
1 \\
1 \\
1 \\
1 \\
1
\end{bmatrix}.\] This yields that $10$ is an eigenvalue of $B$ and the vector $[1\, 1\, 1\, 1\, 1]^{\trans}$ is an eigenvector corresponding to $10$.

Combining these observations, we see that the matrix $B$ has eigenvalues $0$ and $10$ with (algebraic) multiplicities $4$ and $0$, respectively.

Since $A=B+4I$, the eigenvalues of $A$ are $4$ and $14$ with algebraic multiplicities $4$ and $0$, respectively.
(See Problem Eigenvalues and algebraic/geometric multiplicities of matrix $A+cI$.)

The determinant of $A$ is the product of all the eigenvalues of $A$ (counting multiplicities). Thus we have
\[\det(A)=4^4\cdot 14=3584.\] Click here if solved 33

The post Determinant of Matrix whose Diagonal Entries are 6 and 2 Elsewhere first appeared on Problems in Mathematics.

Find all the eigenvalues and eigenvectors of the matrix
\[A=\begin{bmatrix}
3 & 9 & 9 & 9 \\
9 &3 & 9 & 9 \\
9 & 9 & 3 & 9 \\
9 & 9 & 9 & 3
\end{bmatrix}.\]

(Harvard University, Linear Algebra Final Exam Problem)

Hint.

Instead of computing the characteristic polynomial $p(t)=\det(A-tI)$ of $A$, consider the matrix $B=A+6I$.
Then use the relation between eigenvalues of $A$ and $B$.

See the problem “Eigenvalues and algebraic/geometric multiplicities of matrix $A+cI$” for more about the relation.

Solution.

Let $B=A+6I$, where $I$ is the $4 \times 4$ identity matrix. Then every entry of $B$ is $9$. Consider the vector $\mathbf{v}=[1\, 1\, 1\, 1]^{\trans}$.
Then we have
\begin{align*}
B\mathbf{v}=\begin{bmatrix}
9 & 9 & 9 & 9 \\
9 &9 & 9 & 9 \\
9 & 9 & 9 & 9 \\
9 & 9 & 9 & 9
\end{bmatrix}\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix}=36\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix}=36\mathbf{v}.
\end{align*}

It follows that $36$ is an eigenvalue of $B$ and the vector $\mathbf{v}$ is an associated eigenvector.

We apply the elementary row operations and obtain
\[B\to \begin{bmatrix}
1 & 1 & 1 & 1 \\
0 &0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}.\] Hence the rank of $B$ is $1$, and it follows from the rank-nullity theorem that the nullity is $3$.
Thus, $0$ is an eigenvalue of $B$ and its geometric multiplicity is $3$.

The algebraic multiplicity is always greater than or equal to geometric multiplicity.
But the algebraic multiplicity of $0$ cannot be $4$, since $36$ is another eigenvalue.

As a result, the matrix $B$ has eigenvalues $36$ and $0$ with algebraic multiplicities $1$ and $3$.

The eigenvectors corresponding to $36$ are
\[a\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix},\] where $a$ is any nonzero complex number.
The eigenvector corresponding to $0$ are obtained by solving $B\mathbf{x}=\mathbf{0}$.
The solutions satisfy
\[x_1=-x_2-x_3-x_4.\] Hence the eigenvectors corresponding to $0$ are
\[\mathbf{x}=x_2\begin{bmatrix}
-1 \\
1 \\
0 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
-1 \\
0 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
-1 \\
0 \\
0 \\
1
\end{bmatrix},\] where $(x_2, x_3, x_4)\neq (0,0,0)$.

Since $A=B-6I$, it follows that the matrix $A$ has eigenvalues $30$ and $-6$ with algebraic multiplicity $1$ and $3$. Their associated eigenvectors are exactly the eigenvectors for $B$.
(See Problem “Eigenvalues and algebraic/geometric multiplicities of matrix $A+cI$” for a proof of this fact.)

Namely, the eigenvectors corresponding to $30$ are
\[a\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix},\] where $a$ is any nonzero complex number.
The eigenvectors corresponding to $-6$ are
\[x_2\begin{bmatrix}
-1 \\
1 \\
0 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
-1 \\
0 \\
1 \\
0
\end{bmatrix}+x_4\begin{bmatrix}
-1 \\
0 \\
0 \\
1
\end{bmatrix},\] where $(x_2, x_3, x_4)\neq (0,0,0)$.

The post Eigenvalues and Eigenvectors of Matrix Whose Diagonal Entries are 3 and 9 Elsewhere first appeared on Problems in Mathematics.

Find $A^{10}$, where $A=\begin{bmatrix}
4 & 3 & 0 & 0 \\
3 &-4 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 1 & 1
\end{bmatrix}$.

(Harvard University Exam)

Solution.

Let $B=\begin{bmatrix}
4 & 3\\
3& -4
\end{bmatrix}$ and $C=\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix}$ and write $A=\begin{bmatrix}
B & 0\\
0& C
\end{bmatrix}$ as a block matrix.
Then we have
\[A^{10}=\begin{bmatrix}
B^{10} & 0\\
0& C^{10}
\end{bmatrix}.\]

It remains to calculate $B^{10}$ and $C^{10}$.

We have $B^2=5^2I$, where $I$ is the $2\times 2$ identity matrix. From this, we get $B^{10}=5^{10}I$.

Also, we have $C^2=2C$. Applying this repeatedly, we get $C^{10}=2^9C$.

Therefore we have
\[A^{10}=\begin{bmatrix}
5^{10} & 0 & 0 & 0 \\
0 &5^{10} & 0 & 0 \\
0 & 0 & 2^9 & 2^9 \\
0 & 0 & 2^9 & 2^9
\end{bmatrix}.\] Click here if solved 36

The post Calculate $A^{10}$ for a Given Matrix $A$ first appeared on Problems in Mathematics.

Find a basis for the subspace $W$ of all vectors in $\R^4$ which are perpendicular to the columns of the matrix
\[A=\begin{bmatrix}
11 & 12 & 13 & 14 \\
21 &22 & 23 & 24 \\
31 & 32 & 33 & 34 \\
41 & 42 & 43 & 44
\end{bmatrix}.\]

(Harvard University Exam)

Hint.

1. Show that $W=\calN(A^{\trans})$.
2. Find a basis of $\calN(A^{\trans})$ by reducing the matrix $A^{\trans}$.

Solution.

Let us write $A=[A_1 \, A_2 \, A_3 \, A_4]$, where $A_i$ is the $i$-th column vector of $A$ for $i=1,2,3,4$.
First we claim that a vector $\mathbf{x}\in \R^4$ is perpendicular to all column vectors $A_i$ if and only if $\mathbf{x} \in \calN(A^{\trans})$.
To see this, we compute
\begin{align*}
A^{\trans} \mathbf{x} =\begin{bmatrix}
A_1^{\trans} \\
A_2^{\trans} \\
A_3^{\trans} \\
A_4^{\trans}
\end{bmatrix}\mathbf{x}
=\begin{bmatrix}
A_1^{\trans}\mathbf{x} \\
A_2^{\trans} \mathbf{x}\\
A_3^{\trans} \mathbf{x}\\
A_4^{\trans} \mathbf{x}
\end{bmatrix}.
\end{align*}
From this equality the claim follows immediately.

So we proved that $\calN(A^{\trans}) =W$. From this, we see that $W$ is actually a subspace of $\R^4$.

Thus, we need to find a basis for the null space of the transpose $A^{\trans}$.

We apply elementary row operations to $A^{\trans}$ and obtain a reduced row echelon form
\[A^{\trans} \to \begin{bmatrix}
1 & 0 & -1 & -2 \\
0 &1 & 2 & 3 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}.\] The last two columns correspond to two free variables. Let $s$ and $t$ be free variable.
Then $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix} \in \calN(A^{\trans})$ if and only if $\mathbf{x}$ satisfies
\begin{align*}
x_1 &=s+2t \\
x_2 &=-2s-3t\\
x_3 &=s\\
x_4 &=t,
\end{align*}
equivalently
\begin{align*}
\mathbf{x}=s\begin{bmatrix}
1 \\
-2 \\
1 \\
0
\end{bmatrix}
+t\begin{bmatrix}
2 \\
-3 \\
0 \\
1
\end{bmatrix}.
\end{align*}
Therefore a basis of $W=\calN(A^{\trans})$ is
\[ \begin{bmatrix}
1 \\
-2 \\
1 \\
0
\end{bmatrix} \text{ and } \begin{bmatrix}
2 \\
-3 \\
0 \\
1
\end{bmatrix}.\] Click here if solved 37

The post Find a Basis of the Subspace of All Vectors that are Perpendicular to the Columns of the Matrix first appeared on Problems in Mathematics.