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I am reading the introduce of linear system and eigenvalues. There I read if there is a matrix $A$ and vector $x$, it could find a eigenvalue $\lambda$ such that $$Ax = \lambda x$$

I have a really big diagonal matrix D (nxn) (so I don't have to diagonalize it). I need to compute matrix-vector multiplication $Dx$, so based on the eigenvalue property, can I say that I can retain the same result by multiplying the vector $x$ and a number $\lambda$? However, I try to find the eigenvalue of the matrix $D$ with matlab, it does give me $n$ eigenvalues, so which one should I use to estimate $Dx$? Thanks.

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  • $\begingroup$ It's not true in general that $Dx = \lambda x$ for any $x$ you choose. This is only the case if all of the entries on the diagonal are the same. Furthermore, you don't even need to "find" the eigenvalues of $D$. You know what they are immediately: they are the elements on the diagonal. You can see this from the fact that if $e_j$ is a (canonical) basis vector, $De_j = D_{jj} e_j$, where $D_{jj}$ is the $j$-th term down the diagonal. $\endgroup$ – Cameron Williams Sep 21 '13 at 21:00
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When you talk about eigenvalues and eigenvectors and write $Ax = \lambda x$, what it means is the following: Given the matrix $A$, there are special values $\lambda$'s and special vectors $x$ such that $Ax = \lambda x$, i.e., the $\lambda$'s and $x$'s are not arbitrary, and the statement $Ax = \lambda x$ is not true that for any value $\lambda$ and any vector $x$.

In your case, if you are interested in finding the matrix-vector product $Dx$, where $D$ is a diagonal matrix, then all you need to do is the following: $$\begin{bmatrix}d_1 & 0 & 0 & \cdots & 0\\ 0 & d_2 & 0 & \cdots & 0\\ 0 & 0 & d_3 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots& \vdots\\ 0 & 0 & 0 & \cdots & d_n\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\\ \vdots\\ x_n \end{bmatrix} = \begin{bmatrix}d_1x_1\\d_2x_2\\d_3x_3\\ \vdots\\ d_nx_n \end{bmatrix}$$

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