# Spectral Decomposition of A and B.

I was given the following question in my linear algebra course.

Let $A$ be a symmetric matrix, $c >0$, and $B=cA$, find the relationship between the spectral decompositions of $A$ and $B$.

From what I understand. If $A$ is a symmetric matrix, then $A=A^T$. A symmetric matrix has $n$ eigenvalues and there exist $n$ linearly independent eigenvectors (because of orthogonality) even if the eigenvalues are not distinct. Since $B=cA$ and $A=A^T$, then we can conclude that $B=cA^T$, which would imply that $B$ is also symmetric, meaning it also has a linearly independent eigenbasis.

Focusing on $A$, since it has a linearly independent eigenbasis, we have $A = PD_aP^{-1}$ by Spectral decomposition where $P$ is the eigenbasis and $D_a$ is the diagonal matrix of $A$ eigenvalues $\lambda_i$ \begin{array} d D_a & = & \begin{bmatrix} \lambda_1 & & \\ &\ddots&\\ & & \lambda_i \end{bmatrix} \end{array}

Now since $B=cA$, then we have $B=cPD_aP^{-1}$, which can be rewritten as $B = PD_bP^{-1}$, where

\begin{array} d D_b & = & cD_a & =c\begin{bmatrix} \lambda_1 & & \\ &\ddots&\\ & & \lambda_i \end{bmatrix} & = & \begin{bmatrix} c\lambda_1 & & \\ &\ddots&\\ & & c\lambda_i \end{bmatrix} \end{array}

From this I can conclude that $B$ and $A$ actually have the same linearly independent eigenbasis. Furthermore, the eigenvalues of $B$ are a scalar multiple of the eigenvalues of $A$ by a factor of $c$.

Have I fully describe the relationship between $A$ and $B$?

$\dfrac{\lambda}{c} I- A$ isn't invertible if and only if $\lambda I- cA$ isn't invertible.
Hence, $\lambda\in Sp(cA)$ if and only if $\dfrac{\lambda}{c}\in Sp(A)$
Yes, I would say that you have fully described the relationship between $A$ and $B$.