Question

Let $A \in \mathbb{F}^{p \times q}$ and suppose that $\mathcal{R}_A=\mathbb{F}^p$ and $\mathcal{N}_A \dot{+} \mathcal{Y}=$ $\mathbb{F}^q$ for some proper nonzero subspace $\mathcal{Y}$ of $\mathbb{F}^q$. Show that there exists a pseudoinverse $A^{\circ}$ of $A$ such that $\mathcal{N}_{A^{\circ}}=\{\mathbf{0}\}$ and $\mathcal{R}_{A^{\circ}}=\mathcal{Y}$.

   Let $A \in \mathbb{F}^{p \times q}$ and suppose that $\mathcal{R}_A=\mathbb{F}^p$ and $\mathcal{N}_A \dot{+} \mathcal{Y}=$ $\mathbb{F}^q$ for some proper nonzero subspace $\mathcal{Y}$ of $\mathbb{F}^q$. Show that there exists a pseudoinverse $A^{\circ}$ of $A$ such that $\mathcal{N}_{A^{\circ}}=\{\mathbf{0}\}$ and $\mathcal{R}_{A^{\circ}}=\mathcal{Y}$.

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Linear Algebra in Action

Linear Algebra in Action

Harry Dym 1st Edition

Chapter 11, Problem 6

Instant Answer

Step 1

- $\mathcal{R}_A = \mathbb{F}^p$ means that the range (or column space) of $A$ is the entire $\mathbb{F}^p$. This implies that $A$ is a surjective (onto) linear transformation. - $\mathcal{N}_A \dot{+} \mathcal{Y} = \mathbb{F}^q$ means that the null space of $A$ Show more…

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Let $A \in \mathbb{F}^{p \times q}$ and suppose that $\mathcal{R}_A=\mathbb{F}^p$ and $\mathcal{N}_A \dot{+} \mathcal{Y}=$ $\mathbb{F}^q$ for some proper nonzero subspace $\mathcal{Y}$ of $\mathbb{F}^q$. Show that there exists a pseudoinverse $A^{\circ}$ of $A$ such that $\mathcal{N}_{A^{\circ}}=\{\mathbf{0}\}$ and $\mathcal{R}_{A^{\circ}}=\mathcal{Y}$.

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