Determine Whether Given Matrices are Similar

Linear Algebra Problems and Solutions

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 $B=\begin{bmatrix}
1 & 2\\
-1& 4
\end{bmatrix}$?

 
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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.

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1 Response

  1. 06/13/2017

    […] For a solution together with similar problems, see the post “Determine whether given matrices are similar“. […]

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