Short Answer

Consider a quadratic form q on with symmetric matrix A, with rank A = r. Suppose that A has p positive eigenvalues, if eigenvalues are counted with their multiplicities. Show that there exists an orthogonal basis ${\vec{w}}_{1}, . . . {\vec{w}}_{n} of R^{n}$ such that $q (c_{1} {\vec{w}}_{1} + . . . . + c_{n} {\vec{w}}_{n}) = c_{p}^{2} + . . . . + c_{p}^{2} + . . . . + c_{p}^{2} - c_{p + 1}^{2} - . . . . - c_{r}^{2} .$ . Hint: Modify the approach outlined in and 65.

The diagonalizability of a quadratic form and definiteness property are used here to prove this.

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TABLE OF CONTENTS

Step 1 0f 2: Given information

q(x) is the quadratic form, defined by symmetric matrix $A \in R^{n \times n}$ , that is $q (x) = x^{T} Ax$ with rank (A) = r and it has p positive eigen values.
As rank (A) = r , and (n - r ) eigenvalues are zero and since p of the r nonzero eigen values are positive and (r - p) eigenvalue are negative.
Let $λ_{1}, λ_{2}, . . . . ., λ_{p}$ be the positive eigenvalues and the negative eigenvalues are role="math" localid="1659700981027" $λ_{p + 1}, λ_{p + 2}, . . . . ., λ_{r}$ .
$λ_{r + 1}, λ_{p + 2}, . . . . ., λ_{r}$ are all zero eigenvalues.
As per theorem 8.2.2, we know that for orthonormal eigenbasis $B = \{{\vec{v}}_{1}, {\vec{v}}_{2}, . . . {\vec{v}}_{n}\}$ and corresponding eigenvalue $λ_{1}, λ_{2}, . . ., λ_{p}, λ_{p + 1}, . . . ., λr, λ_{r + 1}, . . . ., λ_{n}$ of A, the above quadratic form is diagonalizable as $q (x) = λ_{1} c_{1}^{2} + λ_{2} c_{2}^{2} + . . . + λ_{p} c_{p}^{2} + λ_{p + 1} c_{p + 1}^{2} + . . . + λ_{r} c_{r}^{2} + λ_{r + 1} c_{r + 1}^{2} + . . . + λ_{n} c_{n}^{2}$ , $∴ q (x) = λ_{1} c_{1}^{2} + λ_{2} c_{2}^{2} + . . . + λ_{p} c_{p}^{2} + λ_{p + 1} c_{p + 1}^{2} + . . . + λ_{r} c_{r}^{2}$ where $λ_{i} = 0 for i = r + 1, . . . n$
Here $c_{i} s$ are the coordinates of x with respect to the eigenbasis B
Let us define $C = \{{\vec{w}}_{1}, {\vec{w}}_{2}, . . . ., {\vec{w}}_{n}\}$ where ${\vec{w}}_{i} = \frac{{\vec{v}}_{i}}{\sqrt{|λ_{i}|}} for i = 1, 2 . . . r and {\vec{w}}_{j} = {\vec{v}}_{j}$ for $j = r + 1, . . ., n$ then, is orthogonal basis of as defined.

Step 2 of 2: Application

$q (c_{1} {\vec{w}}_{1} + c_{2} {\vec{w}}_{2} + . . . + c_{n} {\vec{w}}_{n})$

role="math" localid="1659702408287" $= c_{1} (c_{1} \frac{{\vec{v}}_{1}}{\sqrt{|λ_{1}|}} + c_{2} \frac{{\vec{v}}_{2}}{\sqrt{|λ_{2}|}} + . . . + c_{r} \frac{{\vec{v}}_{r}}{\sqrt{|λ_{r}|}} + c_{r + 1} {\vec{v}}_{r + 1} + . . . + c_{n} {\vec{v}}_{n}) = (\frac{c_{1}}{\sqrt{|λ_{1}|}} {\vec{v}}_{1} + . . . + \frac{c_{p}}{\sqrt{|λ_{p}|}} {\vec{v}}_{p} + \frac{c_{p + 1}}{\sqrt{|λ_{p + 1}|}} v_{p + 1} + \frac{c_{r}}{\sqrt{|λ_{r}|}} {\vec{v}}_{r} + c_{r + 1} v_{r + 1} + c_{n} {\vec{v}}_{n}) = λ_{1} {(\frac{c_{1}}{\sqrt{|λ_{1}|}})}^{2} + . . . + λ_{p} {(\frac{c_{p}}{\sqrt{|λ_{p}|}})}^{2} + λ_{p + 1} {(\frac{c_{p + 1}}{\sqrt{|λ_{p + 1}|}})}^{2} + . . . + λ_{r} {(\frac{c_{r}}{\sqrt{|λ_{r}|}})}^{2} + 0 . c_{r + 1}^{2} + . . . + 0 . c_{n}^{2}$

, where $λ_{1} = 0 for i = r + 1, . . ., n$

$= c_{1}^{2} + c_{2}^{2} + . . . + c_{p}^{2} - c_{p + 1}^{2} - c_{p + 2}^{2} - . . . - c_{r}^{2}, where λ_{i} > 0 for i = 1, . . . . p and λ_{j} < 0 j = p + 1, . . . r$

Result:

It is proved using diagonalizability of a quadratic form and definiteness property.

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