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Expert-verified Found in: Page 403 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # If A is a positive definite ${\mathbit{n}}{\mathbf{×}}{\mathbit{n}}$ matrix, and R is any real ${\mathbit{n}}{\mathbf{×}}{\mathbit{n}}$ matrix, what can you say about the definiteness of the matrix ${{\mathbf{R}}}^{{\mathbf{T}}}{\mathbf{AR}}$ ? For which matrices R is ${{\mathbf{R}}}^{{\mathbf{T}}}{\mathbf{AR}}$ positive definite?

${R}^{T}AR$ will be positive definite for all R with $ker\left(R\right)=\left\{0\right\}.$

See the step by step solution

## Step 1 0f 2: Given information

• Let $\stackrel{\to }{x}\in {\mathrm{ℝ}}^{m}$ .
• Let us consider for symmetric matrix A of order n and any real matrix

$R\in {\mathrm{ℝ}}^{n×m},\phantom{\rule{0ex}{0ex}}{\stackrel{\to }{x}}^{T}{R}^{T}AR\stackrel{\to }{x}=\left({\stackrel{\to }{x}}^{T}{R}^{T}\right)\left(A\right)\left(R\stackrel{\to }{x}\right)\phantom{\rule{0ex}{0ex}}={\left(R\stackrel{\to }{x}\right)}^{T}\left(A\right)\left(R\stackrel{\to }{x}\right)\phantom{\rule{0ex}{0ex}}={\stackrel{\to }{y}}^{T}A\stackrel{\to }{y}$

• Here, $\stackrel{\to }{y}=R\stackrel{\to }{x}$

## Step 2 0f 2: Application

• Now, for $\stackrel{\to }{x}\in ker\left(R\right),R\stackrel{\to }{x}=\stackrel{\to }{0}=\stackrel{\to }{y}$and ${\stackrel{\to }{y}}^{T}A\stackrel{\to }{y}=0$thus . For $\stackrel{\to }{x}\notin ker\left(R\right)$ and $\stackrel{\to }{x}\ne 0,\stackrel{\to }{y}=R\stackrel{\to }{x}\ne \stackrel{\to }{0}$ .
• As A is a positive defining matrix, and therefore ${\stackrel{\to }{y}}^{T}A\stackrel{\to }{y}\ge 0.$
• Thus for $\stackrel{\to }{x}\in {\mathrm{ℝ}}^{m}$ , we get ${\stackrel{\to }{x}}^{T}{R}^{T}AR\stackrel{\to }{x}\ge 0$ where ${R}^{T}AR$ is positive semi-definite for any $R\in {\mathrm{ℝ}}^{n×m}$ and symmetric positive definite matrix A .
• When all $\stackrel{\to }{x}\in {\mathrm{ℝ}}^{m},\stackrel{\to }{x}\ne 0,{\stackrel{\to }{x}}^{T}{R}^{T}AR\stackrel{\to }{x}>0$ that is, no $\stackrel{\to }{x}\ne 0$ is in $ker\left(R\right)$ then ${R}^{T}AR$ will be positive definite and $ker\left(R\right)=\left\{0\right\}$ .
• Thus, for all R with $ker\left(R\right)=\left\{0\right\},{R}^{T}AR$ will be positive definite.

Result:

${R}^{T}AR$ Is positive semi-definite for any R and it is Proved using definiteness of A and properties of kernel. Hence, ${R}^{T}AR$ will be positive definite for all with $ker\left(R\right)=\left\{0\right\}$ ### Want to see more solutions like these? 