Determine Whether Given Matrices are Similar Problem 391

(a) Is the matrix $A=\begin{bmatrix} 1 & 2\\ 0& 3 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 3 & 0\\ 1& 2 \end{bmatrix}$?

(b) Is the matrix $A=\begin{bmatrix} 0 & 1\\ 5& 3 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 1 & 2\\ 4& 3 \end{bmatrix}$?

(c) Is the matrix $A=\begin{bmatrix} -1 & 6\\ -2& 6 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 3 & 0\\ 0& 2 \end{bmatrix}$?

(d) Is the matrix $A=\begin{bmatrix} -1 & 6\\ -2& 6 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 1 & 2\\ -1& 4 \end{bmatrix}$? Add to solve later

Solution.

(a) Is the matrix $A=\begin{bmatrix} 1 & 2\\ 0& 3 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 3 & 0\\ 1& 2 \end{bmatrix}$?

Recall that if $A$ and $B$ are similar, then their determinants are the same.
We compute
\begin{align*}
\det(A)=(1)(3)-(2)(0)=3 \text{ and } \det(B)=(3)(2)-(0)(1)=6.
\end{align*}
Thus, $\det(A)\neq \det(B)$, and hence $A$ and $B$ are not similar.

(b) Is the matrix $A=\begin{bmatrix} 0 & 1\\ 5& 3 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 1 & 2\\ 4& 3 \end{bmatrix}$?

It is straightforward to check that $\det(A)=-5=\det(B)$. Thus determinants does not help here.
We recall that if $A$ and $B$ are similar, then their traces are the same. (See Problem “Similar matrices have the same eigenvalues“.)
We compute
\begin{align*}
\tr(A)=0+3=3 \text{ and } \tr(B)=1+3=4,
\end{align*}
and thus $\tr(A)\neq\tr(B)$.
Hence $A$ and $B$ are not similar.

(c) Is the matrix $A=\begin{bmatrix} -1 & 6\\ -2& 6 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 3 & 0\\ 0& 2 \end{bmatrix}$?

We see that
$\det(A)=6=\det(B) \text{ and } \tr(A)=5=\tr(B).$ Thus, the determinants and traces do not give any information about similarity.
The characteristic polynomial of $A$ is given by
\begin{align*}
p(t)&=\det(A-tI)\\
&=\begin{vmatrix}
-1-t & 6\\
-2& 6-t
\end{vmatrix}\\
&=(-1-t)(6-t)-(6)(-2)\\
&=t^2-5t+6.
\end{align*}
(Note that since we found the determinant and trace of $A$, we could have found the characteristic polynomial from the formula $p(t)=t^2-\tr(A)t+\det(A)$.)

Since $p(t)=(t-2)(t-3)$, the eigenvalue of $A$ are $2$ and $3$.
Since $A$ has two distinct eigenvalues, it is diagonalizable.
That is, there exists a nonsingular matrix $S$ such that
$S^{-1}AS=\begin{bmatrix} 2 & 0\\ 0& 3 \end{bmatrix}=B.$ Thus, $A$ and $B$ are similar.

(d) Is the matrix $A=\begin{bmatrix} -1 & 6\\ -2& 6 \end{bmatrix}$ similar to the matrix $B=\begin{bmatrix} 1 & 2\\ -1& 4 \end{bmatrix}$?

We see that
$\det(A)=6=\det(B) \text{ and } \tr(A)=5=\tr(B).$ It follows from the formula $p(t)=t^2-\tr(A)t+\det(A)$ (or just computing directly) that the characteristic polynomials of $A$ and $B$ are both
$t^2-5t+6=(t-2)(t-3).$ Thus, the eigenvalues of $A$ and $B$ are $2, 3$. Hence both $A$ and $B$ are diagonalizable.
There exist nonsingular matrices $S$ and $P$ such that
$S^{-1}AS=\begin{bmatrix} 2 & 0\\ 0& 3 \end{bmatrix} \text{ and } P^{-1}BP=\begin{bmatrix} 2 & 0\\ 0& 3 \end{bmatrix}.$ So we have $S^{-1}AS=P^{-1}BP$, and hence
$PS^{-1}ASP^{-1}=B.$ Putting $U=SP^{-1}$, we have
$U^{-1}AU=B.$ (Since the product of invertible matrices is invertible, the matrix $U$ is invertible.)
Therefore $A$ and $B$ are similar.

Related Question.

For more problems about similar matrices, check out the following posts: Add to solve later

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1. 06/13/2017

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