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Q27E

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Found in: Page 393

Linear Algebra With Applications

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

Diagonalize the ${\mathbit{n}}{\mathbf{×}}{\mathbit{n}}$ matrix $\left[\begin{array}{cccccc}1& 0& 0& \cdots & 0& 1\\ 0& 1& 0& \cdots & 1& 0\\ 0& 0& 1& \cdots & 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 1& 0& \cdots & 1& 0\\ 1& 0& 0& \cdots & 0& 1\end{array}\right]$ . (All ones along both diagonals, and zeros elsewhere.)

If is even the Eigen basis is

${\stackrel{\to }{\mathrm{e}}}_{1}-{\stackrel{\to }{\mathrm{e}}}_{\mathrm{n}},{\stackrel{\to }{\mathrm{e}}}_{2}-{\stackrel{\to }{\mathrm{e}}}_{\mathrm{n}-1},...,{\stackrel{\to }{\mathrm{e}}}_{\left(\mathrm{n}/2\right)+1,}{\stackrel{\to }{\mathrm{e}}}_{1}+{\stackrel{\to }{\mathrm{e}}}_{\mathrm{n}},+{\stackrel{\to }{\mathrm{e}}}_{2}+{\stackrel{\to }{\mathrm{e}}}_{\mathrm{n}-1},...,{\stackrel{\to }{\mathrm{e}}}_{\mathrm{n}/2}+{\stackrel{\to }{\mathrm{e}}}_{\left(\mathrm{n}/2\right)+1}$

With associated eigenvalues 0(n / 2 times ) and 2(n / 2times ).

See the step by step solution

Step 1: The matrix

• A matrix is a rectangular array or table of numbers, symbols, or expressions that are organised in rows and columns to represent a mathematical object or an attribute of such an item in mathematics.
• For example, is a two-row, three-column matrix

Step 2: Determine the Diagonalize matrix

Consider the $\mathrm{n}×\mathrm{n}$ matrix as:

$\mathrm{A}=\left[\begin{array}{cccccc}1& 0& 0& \cdots & 0& 1\\ 0& 1& 0& \cdots & 1& 0\\ 0& 0& 1& \cdots & 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 1& 0& \cdots & 1& 0\\ 1& 0& 0& \cdots & 0& 1\end{array}\right]$

We can write A as A =${\mathrm{J}}_{\mathrm{n}}+{\mathrm{I}}_{\mathrm{n}}$ , where

localid="1660201143401" ${\mathrm{J}}_{\mathrm{n}}=\left[\begin{array}{cccccc}0& 0& 0& \cdots & 0& 1\\ 0& 0& 0& \cdots & 1& 0\\ 0& 0& 0& \cdots & 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 1& 0& \cdots & 1& 0\\ 1& 0& 0& \cdots & 0& 1\end{array}\right]$

${\mathrm{I}}_{\mathrm{n}}=\left[\begin{array}{cccccc}1& 0& 0& \cdots & 0& 0\\ 0& 1& 0& \cdots & 0& 0\\ 0& 0& 1& \cdots & 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& 0& \cdots & 1& 0\\ 0& 0& 0& \cdots & 0& 1\end{array}\right]$

Now consider the transpose of the matrix as represented below:

${\mathrm{J}}_{\mathrm{n}}^{\mathrm{T}}=\left[\begin{array}{cccccc}0& 0& 0& \cdots & 0& 1\\ 0& 0& 0& \cdots & 1& 0\\ 0& 0& 0& \cdots & 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 1& 0& \cdots & 0& 0\\ 1& 0& 0& \cdots & 0& 0\end{array}\right]⇒{\mathrm{J}}_{\mathrm{n}}^{\mathrm{T}}=\left[\begin{array}{cccccc}0& 0& 0& \cdots & 0& 1\\ 0& 0& 0& \cdots & 1& 0\\ 0& 0& 0& \cdots & 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 1& 0& \cdots & 0& 0\\ 1& 0& 0& \cdots & 0& 0\end{array}\right]=\mathrm{J}$

We can see that the matrix ${\mathrm{J}}_{\mathrm{n}}$ is symmetric since ${\mathrm{J}}_{\mathrm{n}}={\mathrm{J}}_{\mathrm{n}}^{\mathrm{T}}$ .

Now evaluate ${\mathrm{J}}_{\mathrm{n}}{\mathrm{J}}_{\mathrm{n}}^{\mathrm{T}}$ as represented below:

${\mathrm{J}}_{\mathrm{n}}{\mathrm{J}}_{\mathrm{n}}^{\mathrm{T}}=\left[\begin{array}{cccccc}0& 0& 0& \cdots & 0& 1\\ 0& 0& 0& \cdots & 1& 0\\ 0& 0& 0& \cdots & 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 1& 0& \cdots & 0& 0\\ 1& 0& 0& \cdots & 0& 0\end{array}\right]\left[\begin{array}{cccccc}0& 0& 0& \cdots & 0& 1\\ 0& 0& 0& \cdots & 1& 0\\ 0& 0& 0& \cdots & 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 1& 0& \cdots & 0& 0\\ 1& 0& 0& \cdots & 0& 0\end{array}\right]$

${\mathrm{J}}_{\mathrm{n}}{\mathrm{J}}_{\mathrm{n}}^{\mathrm{T}}=\left[\begin{array}{cccccc}1& 0& 0& \cdots & 0& 0\\ 0& 1& 0& \cdots & 0& 0\\ 0& 0& 1& \cdots & 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& 0& \cdots & 1& 0\\ 0& 0& 0& \cdots & 0& 1\end{array}\right]={\mathrm{I}}_{\mathrm{n}}$

The matrix ${\mathrm{J}}_{\mathrm{n}}$ is orthogonal since ${\mathrm{J}}_{\mathrm{n}}{\mathrm{J}}_{\mathrm{n}}^{\mathrm{T}}={\mathrm{J}}_{\mathrm{n}}{\mathrm{J}}_{\mathrm{n}}^{\mathrm{T}}={\mathrm{I}}_{\mathrm{n}}$.

In general, if a matrix is symmetric, it is orthogonally diagonalizable and all its eigenvalues are real.

If it is also orthogonal, its eigenvalues must be 1 or -1 .

For a matrix, if n is even, then both eigen-values will be of multiplicity $\frac{\mathrm{n}}{2}$ and if is odd, then the eigenvalue 1 will be of multiplicity $\frac{\mathrm{n}+1}{2}$ and eigen-value -1 will be of multiplicity $\frac{\mathrm{n}-1}{2}$ .

The Eigen value of ${\mathrm{J}}_{\mathrm{n}}$ will be +1 or -1 and when added with Eigen value of ${\mathrm{I}}_{\mathrm{n}}$ which is +1, therefore the eigen values are 0 and 2 with multiplicities $\frac{\mathrm{n}}{2}$ .

The Eigen basis for the Eigen value is:

role="math" localid="1660204574030" ${\stackrel{\to }{\mathrm{e}}}_{1}+{\stackrel{\to }{\mathrm{e}}}_{\mathrm{n}+1};\mathrm{i}=1,...,\frac{\mathrm{n}}{2}$ And for the Eigen value 0 is :

${\stackrel{\to }{\mathrm{e}}}_{1}-{\stackrel{\to }{\mathrm{e}}}_{\mathrm{n}+1};\mathrm{i}=1,...,\frac{\mathrm{n}}{2}$

If n is even the Eigen basis is as represented below:

${\stackrel{\to }{\mathrm{e}}}_{1}-{\stackrel{\to }{\mathrm{e}}}_{\mathrm{n}},{\stackrel{\to }{\mathrm{e}}}_{2}-{\stackrel{\to }{\mathrm{e}}}_{\mathrm{n}-1},...,{\stackrel{\to }{\mathrm{e}}}_{\left(\mathrm{n}/2\right)+1,}{\stackrel{\to }{\mathrm{e}}}_{1}+{\stackrel{\to }{\mathrm{e}}}_{\mathrm{n}},+{\stackrel{\to }{\mathrm{e}}}_{2}+{\stackrel{\to }{\mathrm{e}}}_{\mathrm{n}-1},...,{\stackrel{\to }{\mathrm{e}}}_{\mathrm{n}/2}+{\stackrel{\to }{\mathrm{e}}}_{\left(\mathrm{n}/2\right)+1}$ With associated eigenvalues 0( n/2 times) and 2(n/ 2times) .

Therefore, is even the Eigen basis is

${\stackrel{\to }{\mathrm{e}}}_{1}-{\stackrel{\to }{\mathrm{e}}}_{\mathrm{n}},{\stackrel{\to }{\mathrm{e}}}_{2}-{\stackrel{\to }{\mathrm{e}}}_{\mathrm{n}-1},...,{\stackrel{\to }{\mathrm{e}}}_{\left(\mathrm{n}/2\right)+1,}{\stackrel{\to }{\mathrm{e}}}_{1}+{\stackrel{\to }{\mathrm{e}}}_{\mathrm{n}},+{\stackrel{\to }{\mathrm{e}}}_{2}+{\stackrel{\to }{\mathrm{e}}}_{\mathrm{n}-1},...,{\stackrel{\to }{\mathrm{e}}}_{\mathrm{n}/2}+{\stackrel{\to }{\mathrm{e}}}_{\left(\mathrm{n}/2\right)+1}$

With associated eigenvalues 0 (n/ 2times) and 2(n / 2times) .