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Show that $A_{n\times n}=[a_{jk}]$ is similar to $B_{n\times n}=[(-1)^{j+k}a_{jk}]$

I tried to find an $X$ such that $X^{-1}AX=B$. This is easy for a given 2x2 matrix. But how can I find an X for nxn matrices?

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    @daw its a typo. I meant j+k. – 2017-02-14
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    I have a very different hint/answer. – 2017-02-14

4 Answers 4

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Hint: $$ b_{jk}= ( (-1)^j\delta_{jj}) \cdot a_{jk} \cdot ((-1)^k\delta_{kk}) = \sum_{i,l} ( (-1)^j\delta_{ji}) \cdot a_{il} \cdot ((-1)^k\delta_{lk}) $$ with $\delta_{ij}=0$ if $i\ne j$, $\delta_{ii}=1$.

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Since similarity is an equivalence relation, try exploiting transitivity and using an induction argument. Informally, B is obtained from A by flipping the sign on (some) of the entries. Suppose $B_1$ were obtained from A by flipping only one of (those) entries, could you find the matrix X then (maybe by considering elementary matrix operations)? Then $B_2$ obtained from $B_1$ by flipping one more of (those) entries... etc.

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A hint that is almost an answer...

Let us take the case $n=3$, from which it is easy to establish the general case:

$$\pmatrix{1&&\\&-1&\\&&1}\pmatrix{a&b&c\\d&e&f\\g&h&i}\pmatrix{1&&\\&-1&\\&&1}=\left(\begin{array}{rrr}a&-b&c\\-d&e&-f\\g&-h&i\end{array}\right)$$

(note that this diagonal matrix is its own inverse)

Explanation: Left multiplication (resp. right multiplication) of $\pmatrix{a&b&c\\d&e&f\\g&h&i}$ by $\pmatrix{1&&\\&-1&\\&&1}$ gives a sign change on even numbered columns (resp. on even numbered rows).

Underweaving the two actions provides the desired checkerboard pattern.

Remark: See the use of a similar technique in my answer to (How to classify all nonsingular $n \times n$ matrices $A$ whose all entries in $A$ and $A^{-1}$ are non-negatives.)

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    Well, if the statement is true they must have. The question is how to get from here to there. – 2017-02-14
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    Concerning your rewrite. You're assuming they're similar, plus assuming they're diagonalizable. The problem is given the form of B, to *prove* they are similar. What you wrote is exactly what I wrote on another post, correct given similar diagonalizable matrices. But for this question, that's what you want to prove, so it's a circular argument. – 2017-02-14
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    @victoria In fact, as you can see in this completely different answer, there is an evident X. Btw, I agree with you that my previous argument was seriously uncomplete. – 2017-02-14
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$$det(X) = \sum_{k = 1}^{n!} ( \prod_{r=1}^n x_{r,p_k(r)})$$ where $p_k(r) runs through all the n! permutations of (1,2,3,...,n).

Each of these products takes one element from each row and each column of the matrix X.

Matrix B is the same as A with each entry entry multiplied by its position the "checkerboard" sign matrix

$$\pmatrix{+&-&+&-&\ldots& -\\-&+&-&+ &\ldots & +\\ +&-&+&-&\ldots& -\\-&+&-&+& \ldots&+\\ \ &\ & \ &\ddots\\-&+&-&+& \ldots & +\\}$$

This matrix illustrates concretely $sgn(-1)^{i+j}$ It is helpful in visualizing determinants by cofactors etc., and here to visualize how B is found from A.

We need to figure out the product of the parities $ (-1)^{r + p(r)}$ for A and B and show they are the same, or else the hypothesis is false.

I'll leave you with this much here and see if I can come back later.