Let $S$ be a diagonalizable matrix and $S+5T=I$. Then prove that $T$ is also diagonalizable.

Question

My solution:

Since $S$ is diagonalizable, so we can write $S=P^{-1}DP$, where $P$ is an invertible matrix and $D$ is a diagonal matrix.

Now $5T=I-S=P^{-1}P-P^{-1}DP=P^{-1}(I-D)P$. So $T=P^{-1}\frac{1}{5}(I-D)P$. Since $I-D$ is also a diagonal matrix, hence $T$ is diagonalizable.

Is my proof correct? Can it be done in another way? thanks.

Answer 1

This proof is great!

Here's another way.

Recall that a linear transformation $F:V\to V$ is diagonalizable if and only if there exists a basis of $V$ consisting of eigenvectors of $F$.

In your case, we have two linear transformations $S,T:V\to V$ such that $$ T=\alpha(I-S) $$ where $\alpha=1/5$. Furthermore, $S$ is diagonalizable so there exists a basis $\{s_1,\dotsc,s_n\}$ of $V$ consisting of eigenvectors of $S$. If $\{\lambda_1,\dotsc,\lambda_n\}$ are the corresponding eigenvalues of $S$, then $$ Ts_k=\alpha(I-S)s_k=\alpha(s_k-Ss_k)=\alpha(s_k-\lambda_ks_k)=\alpha(1-\lambda_k)s_k $$ This proves that each $s_k$ is also an eigenvector of $T$ with corresponding eigenvalue $\alpha(1-\lambda_k)$. That is, $\{s_1,\dotsc,s_n\}$ is a basis of $V$ consisting of eigenvectors of $T$. Hence $T$ is diagonalizable.

Answer 2

If a matrix $A$ is diagonalizable, then $\alpha A$ is, for every scalar $\alpha$ (easy). Thus you just need to show that $5T$ is diagonalizable. Since $$ P(5T)P^{-1}=P(I-S)P^{-1}=I-PSP^{-1}=I-D $$ we have that $5T$ is similar to a diagonal matrix.