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Tagged: similar matrix

Linear Algebra

Eigenvalues of Similarity Transformations

Problem 452

Let $A$ be an $n\times n$ complex matrix.
Let $S$ be an invertible matrix.

(a) If $SAS^{-1}=\lambda A$ for some complex number $\lambda$, then prove that either $\lambda^n=1$ or $A$ is a singular matrix.

(b) If $n$ is odd and $SAS^{-1}=-A$, then prove that $0$ is an eigenvalue of $A$.

(c) Suppose that all the eigenvalues of $A$ are integers and $\det(A) > 0$. If $n$ is odd and $SAS^{-1}=A^{-1}$, then prove that $1$ is an eigenvalue of $A$.

 
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Linear Algebra

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

 
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Linear Algebra

Problems and Solutions About Similar Matrices

Problem 319

Let $A, B$, and $C$ be $n \times n$ matrices and $I$ be the $n\times n$ identity matrix.
Prove the following statements.

(a) If $A$ is similar to $B$, then $B$ is similar to $A$.

(b) $A$ is similar to itself.

(c) If $A$ is similar to $B$ and $B$ is similar to $C$, then $A$ is similar to $C$.

(d) If $A$ is similar to the identity matrix $I$, then $A=I$.

(e) If $A$ or $B$ is nonsingular, then $AB$ is similar to $BA$.

(f) If $A$ is similar to $B$, then $A^k$ is similar to $B^k$ for any positive integer $k$.

 
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Linear Algebra

Powers of a Diagonal Matrix

Problem 7

Let $A=\begin{bmatrix}
a & 0\\
0& b
\end{bmatrix}$.
Show that

(1) $A^n=\begin{bmatrix}
a^n & 0\\
0& b^n
\end{bmatrix}$ for any $n \in \N$.

(2) Let $B=S^{-1}AS$, where $S$ be an invertible $2 \times 2$ matrix.
Show that $B^n=S^{-1}A^n S$ for any $n \in \N$

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