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I am studying Linear Algebra. I have faced a problem to understand Symmetric Matrix with spectral decomposition. After I studied spectral decomposition, the next page in my book talks about a positive definite matrix and quadratic form.

I am kind of lost what relationships are there between symmetric decomposition, a positive definite matrix, and quadratic form.

Hope I can have some explanations. Thank you in advance.

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The connection is that every quadratic form can be written in the form $$ Q(x) = x^TAx $$ for the correct real symmetric matrix $A$. For example, for quadratic forms on $2$ variables, we can write $$ Q(x_1,x_2) = ax_1^2 + bx_1x_2 + cx_2^2 = \\ \pmatrix{x_1&x_2} \pmatrix{a&b/2\\b/2&c} \pmatrix{x_1\\x_2} $$ Spectral decompositions give us a "change of variables" that make quadratic forms easier to understand.

In particular, suppose that we have $Q(x) = x^TAx$. $A$ has spectral decomposition $A = UDU^T$ for orthogonal matrix $U$ and diagonal matrix $D$ whose diagonal entries are $\lambda_1,\dots,\lambda_n$. We now have $$ Q(x) = x^T U DU^Tx = (U^Tx)^T D (U^Tx) $$ that is, if we make the substitution $y = U^Tx$ (which is to say $x = Uy$) and define the simpler quadratic form $Q'(x) = x^TDx$, we have $$ Q(x) = Q'(y) = y^TDy = \lambda_1 y_1^2 + \cdots + \lambda_n y_n^2 $$ For example, an important consequence we can gather is Rayleigh's theorem, which says that if $\lambda_1$ is the lowest eigenvalue, then the expression $Q(x)/\|x\|^2$ is minimized when $y_2,\dots,y_{n} = 0$.

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  • $\begingroup$ Here,what does the term, a positive definite matrix, mean? $\endgroup$ – user122358 Apr 9 '15 at 2:36
  • $\begingroup$ Positive definite means it's symmetric and the eigenvalues $\lambda_i$ are positive. $\endgroup$ – Omnomnomnom Apr 9 '15 at 2:46
  • $\begingroup$ Does that mean symmetric matrices always produces positive eigenvalues , therefore they are always positive definite matrices? Or symmetric matrices are positive when their eigenvalues are all positive? $\endgroup$ – user122358 Apr 9 '15 at 2:50
  • $\begingroup$ The second one. $\endgroup$ – Omnomnomnom Apr 9 '15 at 8:31

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