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I know the definition of each one but I don't know how to answers questions about them, or what their properties are and how I can use them to prove/disprove statements about them.

If P, Q, and D are symmetric, anti-symmetric, and diagonal matrices (of the same size) respectively, how would I go about proving if $Q^{2012} + D^{2013} $ is symmetric? Or if $(P + Q)(P - Q)$ is anti-symmetric?

For the first part how do I prove that all square diagonal matrices multiplied by square diagonal matrices are still diagonal? And are anti-symmetric matrices still anti-symmetric if multiplied by themselves?

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    $\begingroup$ You will need to "chase for" definitions of matrices for this problem. Do you know what it means for a matrix to be symmetric? Antisymmetric? Diagonal? $\endgroup$ – NasuSama Jan 12 '14 at 0:07
  • $\begingroup$ Diagonal is only having non-zero values along the leading diagonal, symmetric is when the transpose is the same as the original matrix, and anti-symmetric is when the transpose is the original matrix * -1. I just can't work out in my head how I'd prove or disprove the statements though $\endgroup$ – IBOED2 Jan 12 '14 at 0:13
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    $\begingroup$ To add to NasuSama's point, there are easier definitions to work with than entry-by-entry ones. For example, a symmetric matrix is "symmetric over the diagonal", but an easier definition to work with is $A = A^T$. This way, everything you know about transposes can be brought to bear on the problem. As an example, if $A = A^T$, and $(AB)^T = B^T A^T$, what can you say about $A^n$? $\endgroup$ – Henry Swanson Jan 12 '14 at 0:13
  • $\begingroup$ Here are the definitions: (1) A matrix $A$ is symmetric that satisfies the identity $A = A^{T}$; (2) a matrix $B$ is anti-symmetric that satisfies the identity $B = -B^{T}$, and (3) a matrix $C$ is diagonal if it's a square matrix with the entries outside of the diagonal all zero. From here, we see that all diagonal matrices are symmetric. $\endgroup$ – NasuSama Jan 12 '14 at 0:18
  • $\begingroup$ @HenrySwanson I don't understand how those 2 things relate to each other, what does it say about $A^n$? $\endgroup$ – IBOED2 Jan 12 '14 at 0:33
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In general, given matrices $A,B$ appropriately sized so that $AB$ is defined, we also know that $B^\dagger A^\dagger$ is defined, and in particular that $B^\dagger A^\dagger=(AB)^\dagger.$ (By $\dagger$ I denote transpose.)

It follows that for any square matrix $A$ and any integer $n$ for which $A^n$ is defined (negative $n$ make sense if and only if $A$ is invertible, while nonnegative $n$ always make sense), we have also that $\left(A^\dagger\right)^n$ is defined, and that $\left(A^\dagger\right)^n=(A^n)^\dagger.$ (Why?)

From there, we can readily see that (defined) even powers of antisymmetric matrices are symmetric, as are all (defined) integer powers of symmetric matrices.

Since a sum of symmetric matrices of the same size is again symmetric (why?), then it follows that $Q^{2012}+D^{2013}$ is symmetric. (Why?)

For the second, keep in mind that for any matrix $A$ and any constant $c,$ we have $\left(cA\right)^\dagger=cA^\dagger.$ This, together with the above observations, will allow us to conclude (after some manipulation) that $(P+Q)(P-Q)$ is symmetric.

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It should be observed at the start that the conditions on the matrices $P$, $Q$, and $D$, that they are respectively symmetric, anti-symmetric, and diagonal, implies they are all square. Since they are given to be all the same size, the operations of matrix addition and multiplication between them are defined and permissible.

If $Q$ is an anti- or skew-symmetric matrix, $Q^T = -Q$, whence $Q^2 = -Q^TQ$; for square matrices $A$, we have $(A^TA)^T = A^T(A^T)^T =A^TA$, hence $A^TA$ is symmetric for any square matrix $A$. From this we see that $Q^2$ is symmetric, hence $Q^{2m} = (Q^2)^m$ is also symmetric for any positive integer $m$; this follows from the elementary fact that $(A^T)^m = (A^m)^T$ which is easily seen by a simple application of the two-matrix rule $(AB)^T = B^TA^T$. We also incidentally see that $Q^{2m + 1}$ is skew, $(Q^{2m + 1})^T = -Q^{2m + 1}$: $(Q^{2m + 1})^T = (Q^T)^{2m + 1} = (-Q)^{2m + 1} = (-Q)Q^{2m} = -Q^{2m + 1}$. So, for a skew-symmetric matrix, the even powers are symmetric and the odd powers are skew.

To paraphrase Tom Jefferson, We hold this truth to be (nearly) self-evident: that $D^m$ is diagonal for any diagonal matrix $D$. Nearly self-evident though this assertion may be, it can be easily validated by showing that the product of two diagonal matrices is diagonal, which itself follows from a straightforward application of the definition of matrix multiplication. Since a diagonal matrix is clearly symmetric, we see that for any non-negative integers $m, n$, $Q^{2m + 1} + D^n$, being the sum of two symmetric matrices, is symmetric. In particular, $Q^{2012} + D^{2013}$ is symmetric.

It is a bit harder to see that $(P + Q)(P - Q)$ is skew, because it is in fact symmetric. This follows easily from the symmetry of $A^TA$ for any square $A$: we have $(P - Q)^T = P^T - Q^T = P + Q$ whence $(P + Q)(P - Q) = (P - Q)^T(P - Q)$ is indeed symmetric.

And so it goes . . .

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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