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Found in: Page 245

### Linear Algebra With Applications

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

# Let A be an ${\mathbf{n}}{\mathbf{×}}{\mathbf{m}}$ matrix. Is the formula ${\left(\mathrm{kerA}\right)}^{{\mathbf{\perp }}}{\mathbf{=}}{\mathbf{im}}\left({A}^{T}\right)$ necessarily true? Explain.

Yes.

See the step by step solution

## Step 1: Determine the property of orthogonal.

Consider a $n×m$ matrix A.

Theorem: For any matrix B, ${\mathrm{im}}{\left(\mathrm{B}\right)}^{{\perp }}{=}{{\mathrm{kerB}}}^{{\mathrm{T}}}$ .

By the definition, $\mathrm{im}{\left({\mathrm{A}}^{\mathrm{T}}\right)}^{\perp }={\mathrm{kerA}}^{{\mathrm{T}}^{\mathrm{T}}}$.

As role="math" localid="1659498837555" ${A}^{{T}^{T}}=A$and ${A}^{{\perp }^{\perp }}=A$ , take $\perp$both side in the equation $im{\left({A}^{T}\right)}^{\perp }=ker{A}^{{T}^{T}}$as follows.

$im{\left({A}^{T}\right)}^{\perp }=ker{A}^{{T}^{T}}\phantom{\rule{0ex}{0ex}}im{\left({A}^{T}\right)}^{\perp }=kerA\phantom{\rule{0ex}{0ex}}{\left\{im{\left({A}^{T}\right)}^{\perp }\right\}}^{\perp }={\left(kerA\right)}^{\perp }\phantom{\rule{0ex}{0ex}}im\left({A}^{T}\right)={\left(kerA\right)}^{\perp }$

Hence, the statement is true