Let $A$ be the following $3\times 3$ upper triangular matrix.
\[A=\begin{bmatrix}
1 & x & y \\
0 &1 &z \\
0 & 0 & 1
\end{bmatrix},\] where $x, y, z$ are some real numbers.

Determine whether the matrix $A$ is invertible or not. If it is invertible, then find the inverse matrix $A^{-1}$.

Solution.

We form the augmented matrix
\[[A\mid I]= \left[\begin{array}{rrr|rrr}
1 & x & y & 1 &0 & 0 \\
0 & 1 & z & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 & 1 \\
\end{array} \right]\] and apply elementary row operations as follows.
\begin{align*}
\left[\begin{array}{rrr|rrr}
1 & x & y & 1 &0 & 0 \\
0 & 1 & z & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{R_1-xR_2}
\left[\begin{array}{rrr|rrr}
1 & 0 & y-xz & 1 & -x & 0 \\
0 & 1 & z & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 & 1 \\
\end{array} \right]\\[10pt] \xrightarrow{\substack{R_1-(y-xz)R_3\\ R_2-zR_3}}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & 1 & -x & xz-y \\
0 & 1 & 0 & 0 & 1 & -z \\
0 & 0 & 1 & 0 & 0 & 1 \\
\end{array} \right].
\end{align*}
We could reduce the matrix $A$ into the identity matrix $I$.
Thus, the matrix $A$ is invertible and the right $3\times 3$ matrix is the inverse matrix of $A^{-1}$.
Hence,
\[A^{-1}=\begin{bmatrix}
1 & -x & xz-y \\
0 & 1 & -z \\
0 & 0 & 1
\end{bmatrix}.\] Click here if solved 38

The post The Inverse Matrix of an Upper Triangular Matrix with Variables first appeared on Problems in Mathematics.

Suppose that $A$ and $P$ are $3 \times 3$ matrices and $P$ is invertible matrix.
If
\[P^{-1}AP=\begin{bmatrix}
1 & 2 & 3 \\
0 &4 &5 \\
0 & 0 & 6
\end{bmatrix},\] then find all the eigenvalues of the matrix $A^2$.
 

We give two proofs. The first version is a short proof and uses some facts without proving.
The second proof explains more details and give proofs of the facts which are not proved in the first proof.

Proof (short version).

Let $B=P^{-1}AP$. Since $B$ is an upper triangular matrix, its eigenvalues are diagonal entries $1, 4, 6$.

Since $A$ and $B=P^{-1}AP$ have the same eigenvalues, the eigenvalues of $A$ are $1, 4, 6$. Note that these are all the eigenvalues of $A$ since $A$ is a $3\times 3$ matrix.

It follows that all the eigenvalues of $A^2$ are $1, 4^2, 6^2$, that is, $1, 16, 36$.

Proof (long version.)

Let us put $B:=P^{-1}AP$.
The eigenvalues of $B$ are $1, 4, 6$ since $B$ is an upper triangular matrix and eigenvalues of an upper triangular matrix are diagonal entries.
We claim that the eigenvalues of $A$ and $B$ are the same.

To prove this claim, we show that their characteristic polynomials are equal.
Let $p_A(t), p_B(t)$ be the characteristic polynomials of $A, B$, respectively.
Then we have
\begin{align*}
p_B(t)&=\det(B-tI)=\det(P^{-1}AP-tI)\\
&= \det(P^{-1}(A-tI)P)\\
&=\det(P^{-1})\det(A-tI)\det(P)\\
&=\det(P)^{-1}p_A(t)\det(P)\\
&=p_A(t).
\end{align*}

Therefore, the characteristic polynomials of $A$ and $B$ are the same, and hence the matrices $A$ and $B$ have the same eigenvalues.
Thus $1, 4, 6$ are eigenvalues of the matrix $A$ and these are all the eigenvalues of $A$ since a $3\times 3$ matrix has at most three eigenvalues.

Now we claim that in general if $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ is an eigenvalue of $A^2$. We prove this statement below. Assuming this claim for the moment, we finish the problem.
By the claim, the matrix $A^2$ has eigenvalues $1^2, 4^2, 6^2$. Thus all the eigenvalues of $A^2$ are
\[1, 16, 36.\]

To complete the solution, let us prove the last claim.
Since $\lambda$ is an eigenvalue of $A$, we have an eigenvector $\mathbf{x}$ such that
\[A\mathbf{x}=\lambda \mathbf{x}.\]

Multiplying by $A$ we obtain
\begin{align*}
A^2T\mathbf{x}&=\lambda (A\mathbf{x})\\
&=\lambda (\lambda \mathbf{x})\\
&=\lambda^2 \mathbf{x}
\end{align*}
and thus $\lambda^2$ is an eigenvalue of $A^2$.

The post Eigenvalues of Squared Matrix and Upper Triangular Matrix first appeared on Problems in Mathematics.

Find a square root of the matrix
\[A=\begin{bmatrix}
1 & 3 & -3 \\
0 &4 &5 \\
0 & 0 & 9
\end{bmatrix}.\]

How many square roots does this matrix have?

(University of California, Berkeley Qualifying Exam)
 

Proof.

We will find all matrices $B$ such that $B^2=A$. Such matrices $B$ are square roots of the matrix $A$.

Note that since $A$ is a diagonal matrix, the eigenvalues of $A$ are diagonal entries $1, 4, 9$. Since $A$ has three distinct eigenvalues, it is diagonalizable.
Solving $(A-\lambda I)\mathbf{x}=\mathbf{0}$ for $\lambda=1,4,9$, we find eigenvectors corresponding to eigenvalues $1, 4, 9$ are respectively

\[ \begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}, \quad
\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix} , \quad
\begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix}.\]

Thus the invertible matrix
\[P=\begin{bmatrix}
1 & 1 & 0 \\
0 &1 &1 \\
0 & 0 & 1
\end{bmatrix}\] diagonalizes the matrix $A$, that is, we have
\[P^{-1} AP=\begin{bmatrix}
1 & 0 & 0 \\
0 &4 &0 \\
0 & 0 & 9
\end{bmatrix}.\]

Then if $B^2=A$, then we have $(P^{-1}BP)(P^{-1}B)=P^{-1}AP$.
Let $A’=P^{-1}AP$ and $B’=P^{-1}BP$.

Since we have $B’^2=A’$, we have $B’A’=B’^3=A’B’$.
Since $A’$ is diagonal with distinct diagonal entries, this implies that $B’$ is also a diagonal matrix.

A diagonal matrix $B’$ satisfying $B’^2=A’=\begin{bmatrix}
1 & 0 & 0 \\
0 &4 &0 \\
0 & 0 & 9
\end{bmatrix}$ is one of
\[\begin{bmatrix}
\pm 1 & 0 & 0 \\
0 &\pm 2 &0 \\
0 & 0 & \pm 3
\end{bmatrix}.\] Hence $B$ must be one of
\[P\begin{bmatrix}
\pm 1 & 0 & 0 \\
0 &\pm 2 &0 \\
0 & 0 & \pm 3
\end{bmatrix}P^{-1}.\] The inverse matrix of $P$ can be calculated as
\[P^{-1}=\begin{bmatrix}
1 & -1 & 1 \\
0 &1 &-1 \\
0 & 0 & 1
\end{bmatrix}.\] Therefore, all the square roots of the matrix $A$ are
\[\begin{bmatrix}
1 & 1 & 0 \\
0 &1 &1 \\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
\pm 1 & 0 & 0 \\
0 &\pm 2 &0 \\
0 & 0 & \pm 3
\end{bmatrix}\begin{bmatrix}
1 & -1 & 1 \\
0 &1 &-1 \\
0 & 0 & 1
\end{bmatrix}\] and we have $8$ square root matrices.

For example, when the diagonal matrix has all positive entries, then one of the square roots is
\[\begin{bmatrix}
1 & 1 & 0 \\
0 &1 &1 \\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
1 & 0 & 0 \\
0 & 2 &0 \\
0 & 0 & 3
\end{bmatrix}\begin{bmatrix}
1 & -1 & 1 \\
0 &1 &-1 \\
0 & 0 & 1
\end{bmatrix}=\begin{bmatrix}
1 & 1 & -1 \\
0 &2 &1 \\
0 & 0 & 3
\end{bmatrix}.\]

Related Question.

For a solution of this problem, see the post
A Positive Definite Matrix Has a Unique Positive Definite Square Root

The post Square Root of an Upper Triangular Matrix. How Many Square Roots Exist? first appeared on Problems in Mathematics.

Calculate the determinants of the following $n\times n$ matrices.
\[A=\begin{bmatrix}
1 & 0 & 0 & \dots & 0 & 0 &1 \\
1 & 1 & 0 & \dots & 0 & 0 & 0 \\
0 & 1 & 1 & \dots & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \dots & \dots & \ddots & \vdots \\
0 & 0 & 0 &\dots & 1 & 1 & 0\\
0 & 0 & 0 &\dots & 0 & 1 & 1
\end{bmatrix}\]

The entries of $A$ is $1$ at the diagonal entries, entries below the diagonal, and $(1, n)$-entry.
The other entries are zero.
\[B=\begin{bmatrix}
1 & 0 & 0 & \dots & 0 & 0 & -1 \\
-1 & 1 & 0 & \dots & 0 & 0 & 0 \\
0 & -1 & 1 & \dots & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \dots & \dots & \ddots & \vdots \\
0 & 0 & 0 &\dots & -1 & 1 & 0\\
0 & 0 & 0 &\dots & 0 & -1 & 1
\end{bmatrix}.\]

The entries of $B$ is $1$ at the diagonal entries.
The entries below the diagonal and $(1,n)$-entry are $-1$.
The other entries are zero.

Hint.

1. Calculate the first row cofactor expansion.
2. The determinant of a triangular matrix is the product of its diagonal entries.

Solution.

Apply the cofactor expansion corresponding to the first row. We obtain
\begin{align*}
\det(A)&=
\begin{vmatrix}
1 & 0 & \dots & 0 & 0 & 0 \\
1 & 1 & \dots & 0 & 0 & 0 \\
\vdots & \vdots & \dots & \ddots & \vdots & \vdots \\
0 & 0 &\dots & 1 & 1 & 0\\
0 & 0 &\dots & 0 & 1 & 1
\end{vmatrix}
+(-1)^{n+1}
\begin{vmatrix}
1 & 1 & 0 & \dots & 0 & 0 \\
0 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 &\dots & 1 & 1 \\
0 & 0 & 0 &\dots & 0 & 1
\end{vmatrix}
\end{align*}
The two smaller (minor) $n-1 \times n-1$ matrices are both triangular.
The determinant of a triangular matrix is the product of its diagonal entries.
Thus we see that
\begin{align*}
\det(A)&=1+(-1)^{n+1} \\
&= \begin{cases}
2 & \text{ if } n \text{ is odd}\\
0 & \text{ if } n \text{ is even}.
\end{cases}
\end{align*}
Next we calculate $\det(B)$. By the first row cofactor expansion , we obtain
\begin{align*}
\det(B)&=\\
&\begin{vmatrix}
1 & 0 & \dots & 0 & 0 & 0 \\
-1 & 1 & \dots & 0 & 0 & 0 \\
\vdots & \vdots & \dots & \ddots & \vdots & \vdots \\
0 & 0 &\dots & -1 & 1 & 0\\
0 & 0 &\dots & 0 & -1 & 1
\end{vmatrix}
+(-1)^{n+1}(-1)
\begin{vmatrix}
-1 & 1 & 0 & \dots & 0 & 0 \\
0 & -1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 &\dots & -1 & 1 \\
0 & 0 & 0 &\dots & 0 & -1
\end{vmatrix}.
\end{align*}
The two minor matrices are both triangular.
All the diagonal entries of the first minor matrix are $1$ and those of the second minor matrix are $-1$.
Thus we have
\begin{align*}
\det(B)&=1+(-1)^{n}(-1)^{n-1}=0.
\end{align*}

The post Calculate Determinants of Matrices first appeared on Problems in Mathematics.