Find Jordan Form of αA (α is a scalar, A a matrix)

Question

In my linear algebra course I have a problem which goes as follows:

Suppose A is an nxn matrix over field (R) And J(A) is the jordan form of A.

Given α belongs to field R, what is the jordan form of αA?

I have worked out, through trial and error, that the jordan form of αA is simply J(A) with the elements (the different eigenvalues) along the diagonal multiplied by α.

I couldnt think of a formal proof for this problem, I think it might be some simple characteristic I may have overlooked.

Any help is appreciated Thanks!

Answer 1

$P^{-1}(\alpha A)P=\alpha P^{-1}AP=\alpha J(A)$ which is similar to a Jordan matrix which is the same as $J(A)$ but the eigenvalues multiplied by $\alpha$.

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ok GitGud, take a matrix with $\lambda$ on the diagonal and $\alpha$ on the super diagonal, say the matrix is $J$. Then $P=$Dg$[1,1/\alpha,1/\alpha^2,1/\alpha^3,\ldots]$ is a matrix such that $P^{-1}JP$ is in Jordan form. I hope this is sufficient information for you to continue the proof...(referring to the OP)

Answer 2

Since you are given $J(A)$, Jordan form of $A$, then there must exist an invertible $P$ such that:

$$P^{-1}AP = J(A)$$

where $J(A)$ is a Jordan Matrix and thus is composed of smaller Jordan blocks, $J(x)$ for each eigenvalue, $x$, of A.

For each block in $J(x)$ of size $n \times n$, where $x$ is a eigenvalue, $P$ will have $n$ columns which are 'responsible' for the block $J(x)$ in $J(A)$.

We can then take each $i$'th column for $1 \le i le n$ and multpiply it by $α^{i-1}$.

Do this for each Jordan block in $J(A)$ and you will obtain the desired $P$ for $J(αA) = P^{-1}(αA)P$.