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Consider a matrix $A \in \mathbb{R}^{n \times n}$ and the canonical inner product in $\mathbb{R}^{n}$. Show that if the rows of A form an orthogonal set, the same happens with the columns.

So what I started to say was that if the rows of A form an orthogonal set then:

$ <a_{1,1}, a_{1,2}, ..., a_{1,n}>$ $ <a_{2,1}, a_{2,2}, ..., a_{2,n}>$ ... $ <a_{n,1}, a_{n,2}, ..., a_{n,n}>$ = 0

Now what I'm trying is that if we choose the rows 2 by 2 we get to, for example:

$ <a_{1,1}, a_{1,2}, ..., a_{1,n}>$ $ <a_{2,1}, a_{2,2}, ..., a_{2,n}>$ = $a_{1,1}a_{2,1} + a_{1,2}a_{2,2} + ... + a_{1,n}a_{2,n}$

And now this can manipulate this to get to $ <a_{1,1}, a_{2,1}, ..., a_{n,1}>$ $ <a_{1,2}, a_{2,2}, ..., a_{n,2}>$

But I'm finding hard to manipulate this correctly (maybe it's obvious and I'm not seeing it...

Can someone give a hint on how to proceed?

Martin Argerami
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Granger Obliviate
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1 Answers1

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The question requires that $A$ has no zero row, because otherwise the result is not true. For instance, consider $$ A=\begin{bmatrix}1&1&0\\0&0&1\\0&0&0\end{bmatrix} $$

More than that, the question requires that the rows of $A$ are orthonormal (more properly, that all rows have the same norm). Otherwise, consider $$ A=\begin{bmatrix}1&1\\-2&2\end{bmatrix}, $$ where the rows are orthogonal but the columns are not.

Now for the case where the rows of $A$ are orthonormal (if the all the rows have the same norm $\alpha$, we can multiply $A$ by $\alpha^{-1}$ to make all rows have norm equal to $1$):

The property that the rows of $A$ are orthonormal can be summarized as $AA^T=I$. The desired property, that the columns are orthogonal, can be written as $A^TA=I$.

As $AA^T=I$, we have that $A$ is non-singular (for instance, take determinants) and $A^T=A^{-1}$. But then $A^T$ commutes with $A$ and $$A^TA=I.$$

Martin Argerami
  • 205,756