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Apologies if this has been answered already but I can't seem to find an answer that I think answers my question (or at least one I understand).

Anyways the question is,


    Prove ‖x‖22=xTx for all x in Rn

When I took a swing at it I got a scalar value for the Euclidian (or Frobenius?) norm and a matrix value (of dimension n x n) for x-transpose times x. I really have no clue what I'm doing but I know it can't be good if I have a scalar and matrix set to be equivalent to one another.

thanks

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  • $\begingroup$ Note that $\mathbf{ x}^T$ is ${1}\times n$ and $\mathbf{x}$ is $n\times 1$, so $\mathbf{x}^T\mathbf{x}$ is size $1\times 1$ (essentially a scalar). $\endgroup$ Mar 20, 2019 at 5:11

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If we think of $x \in \mathbb{R}^d$ as a column vector, then $$x^tx = \sum_{j=1}^d x_j^2 = \lvert x \rvert^2.$$ Edit: It seems you may be a bit confused on matrix multiplication. Since $x^t$ is a $1 \times d$ matrix and $x$ is a $d \times 1$ matrix, $x^tx$ will be a $1\times 1$ matrix (ie, a scalar): $$x^tx = (x_1, \ldots, x_d) \begin{pmatrix} x_1 \\ \vdots \\ x_d \end{pmatrix} = \sum_{j=1}^d x_j^2.$$

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  • $\begingroup$ aha thanks, for some reason I thought a matrix in R^n would be a row vector, not a column vector, so when i did the matrix multiplication it came out wonky. $\endgroup$
    – saucin'
    Mar 20, 2019 at 8:56

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