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### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Consider the function ${\mathbf{T}}\left(A\right)\left(\stackrel{⇀}{X}\right){\mathbf{=}}{\stackrel{\mathbf{⇀}}{\mathbf{x}}}^{{\mathbf{T}}}{\mathbf{A}}\stackrel{\mathbf{⇀}}{\mathbf{x}}{\mathbf{}}{\mathbf{from}}{\mathbf{}}{{\mathbf{R}}}^{{\mathbf{nxn}}}{\mathbf{}}{\mathbf{to}}{\mathbf{}}{{\mathbf{Q}}}_{{\mathbf{n}}}$ . Show that T is a linear transformation. Find the image, kernel, rank, and nullity of T.

the solution is

$\mathrm{ker}\left(\mathrm{T}\right)=\left\{\mathrm{A}\in {\mathrm{R}}^{\mathrm{nxn}}:{\mathrm{A}}^{\mathrm{T}}=-\mathrm{A}\right\}\phantom{\rule{0ex}{0ex}}\mathrm{img}\left(\mathrm{T}\right)=\left\{{\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}\in {\mathrm{Q}}_{\mathrm{n}}:\mathrm{A}\in {\mathrm{R}}^{\mathrm{nxn}},\mathrm{A}={\mathrm{A}}^{\mathrm{T}}\right\}\phantom{\rule{0ex}{0ex}}\mathrm{rank}\left(\mathrm{T}\right)=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\phantom{\rule{0ex}{0ex}}\mathrm{nulity}\left(\mathrm{T}\right)=\frac{\mathrm{n}\left(\mathrm{n}-1\right)}{2}$

See the step by step solution

## Step 1: Given information

$\mathrm{T}\left(\mathrm{A}\right)\left(\stackrel{⇀}{\mathrm{x}}\right)=\stackrel{⇀{\mathrm{T}}^{}}{\mathrm{x}}\mathrm{A}\stackrel{⇀}{\mathrm{x}}$

## Step 2: Linear transformation and image of T and kernel of T

Consider $\mathrm{A},\mathrm{B}\in {\mathrm{R}}^{\mathrm{nxn}}\mathrm{and}\mathrm{\alpha }\in \mathrm{R},$then

$\mathrm{T}\left(\mathrm{\alpha A}+\mathrm{B}\right)\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{T}}\left(\mathrm{\alpha A}+\mathrm{B}\right)\mathrm{x}\phantom{\rule{0ex}{0ex}}={\mathrm{x}}^{\mathrm{T}}\left(\mathrm{\alpha A}\right)\mathrm{x}+{\mathrm{x}}^{\mathrm{T}}\mathrm{Bx}\phantom{\rule{0ex}{0ex}}={\mathrm{\alpha x}}^{\mathrm{T}}\mathrm{Ax}+{\mathrm{x}}^{\mathrm{T}}\mathrm{Bx}\phantom{\rule{0ex}{0ex}}=\mathrm{\alpha T}\left(\mathrm{A}\right)\left(\mathrm{x}\right)+\mathrm{T}\left(\mathrm{B}\right)\left(\mathrm{x}\right)$

$\mathrm{For}\mathrm{\alpha }=1,\mathrm{we}\mathrm{get}\mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right)\mathrm{and}\mathrm{for}\mathrm{B}=0,\mathrm{we}\mathrm{get}\mathrm{T}\left(\mathrm{\alpha A}\right)=\mathrm{\alpha T}\left(\mathrm{A}\right).\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{T}:{\mathrm{R}}^{\mathrm{NXN}}\to {\mathrm{Q}}_{\mathrm{n}},\mathrm{T}\left(\mathrm{A}\right)\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{transformation}.\phantom{\rule{0ex}{0ex}}\mathrm{Image}\mathrm{of}\mathrm{T}$

We know that for any symmetric matrix $\mathrm{A}\in {\mathrm{R}}^{\mathrm{nxn}}$times $\mathrm{n},{\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}$, creates a quadratic form(unique), and that the image of T defines the entire space of quadratic form ${\mathrm{Q}}_{\mathrm{n}}$ described by the symmetric matrices.

$\mathrm{rank}\left(\mathrm{T}\right)=\mathrm{dim}\left(\mathrm{lmg}\left(\mathrm{T}\right)\right)=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}$

Kernel of T: if A is a skew symmetric matrix, that is ${\mathrm{A}}^{\mathrm{T}}=-\mathrm{A}$ then ,

$\left({\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}\right)={\mathrm{x}}^{\mathrm{T}}{\mathrm{A}}^{\mathrm{T}}\mathrm{x}=-{\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}\phantom{\rule{0ex}{0ex}}\mathrm{But},\mathrm{since}{\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}\mathrm{is}\mathrm{a}\mathrm{real}\mathrm{number},\mathrm{so}{\left({\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}\right)}^{\mathrm{T}}={\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}\mathrm{and}\mathrm{so},\mathrm{we}\mathrm{get}$

${\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}=-{\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}⇒{\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}=0$. So, the space of all skew-symmetric matrices is the subset of the kernel of T. By rank-nullity theorem, dimension of kernel of T= nullity of T ${\mathrm{dimR}}^{\mathrm{nxn}}={\mathrm{n}}^{2}\mathrm{and}\mathrm{rank}\left(\mathrm{T}\right)=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}$ As a result, because is the dimension of the space of skew-symmetric matrices, all skew-symmetric matrices make up the whole kernel of T.

Thus,

$\mathrm{ker}\left(\mathrm{T}\right)=\left\{\mathrm{A}\in {\mathrm{R}}^{\mathrm{nxn}}:{\mathrm{A}}^{\mathrm{T}}=-\mathrm{A}\right\}\phantom{\rule{0ex}{0ex}}\mathrm{lmg}\left(\mathrm{T}\right)=\left\{{\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}\in {\mathrm{Q}}_{\mathrm{n}}:\mathrm{A}\in {\mathrm{R}}^{\mathrm{nxn}},\mathrm{A}={\mathrm{A}}^{\mathrm{T}}\right\}={\mathrm{Q}}_{\mathrm{n}}$

## Step 3: Conclusion

$\mathrm{ker}\left(\mathrm{T}\right)=\left\{\mathrm{A}\in {\mathrm{R}}^{\mathrm{nxn}}:{\mathrm{A}}^{\mathrm{T}}=-\mathrm{A}\right\}\phantom{\rule{0ex}{0ex}}\mathrm{lmg}\left(\mathrm{T}\right)=\left\{{\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}\in {\mathrm{Q}}_{\mathrm{n}}:\mathrm{A}\in {\mathrm{R}}^{\mathrm{nxn}},\mathrm{A}={\mathrm{A}}^{\mathrm{T}}\right\}\phantom{\rule{0ex}{0ex}}\mathrm{rank}\left(\mathrm{T}\right)=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\phantom{\rule{0ex}{0ex}}\mathrm{nulity}\left(\mathrm{T}\right)=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}$