Transpose matrix and inner product

Question

In one of the proofs in class there was given the equality for the dot product:

$$\langle Ax, Ax\rangle = \langle x, A^tAx\rangle$$

I don't understand why this is correct. Is there a way to show this without explicitly looking at the multiplications and sums?

thanks.

score 21 · Accepted Answer · answered Jun 18 '18 at 14:57

21

This is simpler than you might think.

By definition, $$\langle x, y \rangle = x^{T}y\text{.}$$ Hence, $$\langle Ax, Ax \rangle = (Ax)^{T}Ax$$ but by properties of matrix transposition, recall that $$(Ax)^{T}=x^{T}A^{T}$$ hence $$\langle Ax, Ax \rangle = (Ax)^{T}Ax = x^{T}A^{T}Ax = (x)^{T}(A^TAx)=\langle x, A^TAx \rangle\text{.}$$

answered Jun 18 '18 at 14:57

Clarinetist

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1

Thanks for the nice answer. Just for beginners like me: To be general, $\langle x,y \rangle = x^y$ if it's a complex, where $x^$ denotes a conjugate transpose. – starriet 차주녕 Oct 17 '22 at 01:30

score 3 · Answer 2 · answered Jun 18 '18 at 14:56

3

By definition

$$\langle Ax, Ax\rangle = x^TA^TAx= \langle x, A^TAx\rangle$$

answered Jun 18 '18 at 14:56

user

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Transpose matrix and inner product

2 Answers2