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What does the following mean? I know that $x^Tx$ is the magnitude of $x$. What does the following formula represent intuitively? $x$ is a vector and $A$ is some scaling matrix. The given is $x^T A x i$

(Ignore the i)

I'm learning about positive definiteness.

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  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$
    – Shaun
    Mar 16, 2018 at 1:20
  • $\begingroup$ Note that $x$ and $X$ are two different things. Please do not pretend they are the same. $\endgroup$ Mar 16, 2018 at 1:28
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    $\begingroup$ Wikipedia: Definite quadratic form will be helpful. $x^TAx$ comes from a matrix form of a quadratic form. $\endgroup$
    – ChoF
    Mar 16, 2018 at 2:41
  • $\begingroup$ $x^Tx$ is the square of the magnitude of $x$ (assuming, that is, that the coordinates are relative to the standard basis). $\endgroup$
    – amd
    Mar 17, 2018 at 3:31

3 Answers 3

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Rather than thinking about $y^T x$, it's better to think about the dot product $y \cdot x$. Recall that in a real vector space, the dot product $y \cdot x$ gives you a measure of how the vectors $x$ and $y$ project onto each other: if the dot product is large and positive, the vectors $x$ and $y$ are pointing in a similar direction, and if it is large and negative, the vectors are pointing in an opposite direction.

Since $x^T A x = x \cdot (Ax)$, it is a measure of how "pointing in the same direction" the vector $Ax$ is compared to $x$. The matrix $A$ is called positive definite if $x \cdot (Ax) > 0$ for all $x \neq 0$. This gives some immediate consequences:

  • $A$ has full rank, so is an invertible matrix.
  • For each vector $x$, the vectors $x$ and $Ax$ are always on the same side of the hyperplane perpendicular to $x$.
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  • $\begingroup$ Thank you for this insight. This cleared up a few things for me! $\endgroup$
    – Oiler
    Jan 25, 2021 at 19:55
  • $\begingroup$ The dot approach really helps. Great answer. $\endgroup$
    – anonymus
    Jun 22, 2021 at 15:19
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If this was in one dimension, what would you get?

You would get a second degree polynomial $ax^2$. Imagine now instead dimension $37$ and a function $f(x_1,x_2 ... , x_{37})=x_1^2 + x_2^2 + ... + x_{37}^2$. This is very cumbersome to type, $x^TAx$ is a generalization to higher dimension, a much more compact form to write a second degree polynomial instead of writing all terms out.

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If $X$ is a vector in $n$ dimensional space, it is an $n \times 1$ matrix-a column. $X^T$ is a $1 \times n$ matrix-a row. $A$ is an $n \times n$ matrix, which you might see as the matrix of a linear transformation on your space. $AX$ is the vector that $A$ transforms $X$ into. It is again $n \times 1$. Then $X^TAX$ is a $1 \times 1$ matrix, a number. $A$ is positive definite if the number is positive for all nonzero $X$. An example is the identity matrix, because $IX=X$ and we know $X^TX$ is positive for all nonzero $X$.

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