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Expert-verified Found in: Page 325 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Consider the linear space ${\mathbit{V}}$ of all ${\mathbit{n}}{\mathbf{×}}{\mathbit{n}}$ matrices for which all the vectors ${\stackrel{\mathbf{⇀}}{\mathbf{e}}}_{{\mathbf{1}}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{,}}{\stackrel{\mathbf{⇀}}{\mathbf{e}}}_{{\mathbf{n}}}$ are eigenvectors. Describe the space ${\mathbit{V}}$ (the matrices in ${\mathbit{V}}$ "have a name"), and determine the dimension of ${\mathbit{V}}$.

Hence, the required dimension is n .

See the step by step solution

## Step 1: Definition of the Eigenvectors

Eigenvectors are a nonzero vector that is mapped by a given linear transformation of a vector space onto a vector that is the product of a scalar multiplied by the original vector.

## Step 2: Find dimension

Let’s consider the linear space $V$or all $n×n$ matrices which all the vectors role="math" localid="1659528394250" ${\stackrel{⇀}{\mathrm{e}}}_{\mathrm{b}},{\stackrel{⇀}{\mathrm{e}}}_{\mathrm{n}}$are Eigen vectors.

We want to describe the space and determine its dimension.

The set of matrices in the space $V$ spanned by the Eigen vectors ${\stackrel{⇀}{\mathrm{e}}}_{\mathrm{b}},{\stackrel{⇀}{\mathrm{e}}}_{\mathrm{n}}$are called diagonal matrices.

If you put all the vectors ${\stackrel{⇀}{\mathrm{e}}}_{z},{\stackrel{⇀}{\mathrm{e}}}_{\mathrm{n}}$together in one large $n×n$matrix it will have only values across the diagonal, thus the name diagonal matrix.

Since all $n$ Eigen vectors span the space and are linearly independent.

Therefore, $dimV=n$. ### Want to see more solutions like these? 