# Eigenvalues

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• October 14th 2011, 07:42 PM
dwsmith
Eigenvalues
M is a nxn matrix where all entries are 1's.

My conjecture is that n - 1 eigenvalues are 0 and the nth eigenvalues is one.

If this is correct, how do I go about proving it?
• October 14th 2011, 08:38 PM
Deveno
Re: Eigenvalues
did you test your conjecture for small values of n? for a 2x2 matrix:

$\begin{vmatrix}1-x&1\\1&1-x\end{vmatrix} = (1-x^2) - 1 = 1 - 2x + x^2 - 1 = x^2 - 2x = x(x-2)$

which appears to have eigenvalues 0 and 2.
• October 14th 2011, 08:50 PM
dwsmith
Re: Eigenvalues
Quote:

Originally Posted by Deveno
did you test your conjecture for small values of n? for a 2x2 matrix:

$\begin{vmatrix}1-x&1\\1&1-x\end{vmatrix} = (1-x^2) - 1 = 1 - 2x + x^2 - 1 = x^2 - 2x = x(x-2)$

which appears to have eigenvalues 0 and 2.

Now I am not sure what it may be.
• October 14th 2011, 09:41 PM
Deveno
Re: Eigenvalues
try a 3x3 matrix next, maybe you'll get a theory?
• October 14th 2011, 10:28 PM
FernandoRevilla
Re: Eigenvalues
Using the steps:

$(i)\;R_i\to R_i-R_1\;(i>1)\qquad (ii)\;C_1\to C_1+C_2+\ldots C_n$

we obtain

\displaystyle\begin{aligned}\begin{vmatrix} 1-\lambda & 1 & \ldots & 1\\ 1 &1-\lambda & \ldots & 1 \\ \vdots&&&\vdots \\ 1 & 1 &\ldots & 1-\lambda\end{vmatrix}&=\begin{vmatrix} 1-\lambda & \;\;1 & \ldots & \;\;1\\ \lambda &-\lambda & \ldots & \;\;0 \\ \vdots&&&\vdots \\ \lambda & \;\;0 &\ldots & -\lambda\end{vmatrix}\\&=\begin{vmatrix} n-\lambda & 1 & \ldots & 1\\ 0 &-\lambda & \ldots & 0 \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & -\lambda\end{vmatrix}\\&=(n-\lambda)(-\lambda)^{n-1}\end{aligned}

So, the eigenvalues are $\lambda=0$ (n-1-uple) and $\lambda=n$ (1-uple) if $n\geq 2$ . If $n=1$ we have only the eigenvalue $\lambda=1$ (1-uple).
• October 14th 2011, 11:46 PM
Deveno
Re: Eigenvalues
i would have been inclined use induction and expand along the top row, but that works great!
• October 15th 2011, 01:42 AM
FernandoRevilla
Re: Eigenvalues
In general with the steps

$(i)\;R_i\to R_i-R_1\;(i>1)\qquad (ii)\;C_1\to C_1+C_2+\ldots C_n$

we triangularize the determinant

$\begin{vmatrix} a & b & \ldots & b\\ b & a & \ldots & b \\ \vdots&&&\vdots \\ b & b &\ldots & a\end{vmatrix}=\ldots=[a+(n-1)b](a-b)^{n-1}$
• October 15th 2011, 03:38 PM
dwsmith
Re: Eigenvalues
Quote:

Originally Posted by FernandoRevilla
Using the steps:

$(i)\;R_i\to R_i-R_1\;(i>1)\qquad (ii)\;C_1\to C_1+C_2+\ldots C_n$

we obtain

\displaystyle\begin{aligned}\begin{vmatrix} 1-\lambda & 1 & \ldots & 1\\ 1 &1-\lambda & \ldots & 1 \\ \vdots&&&\vdots \\ 1 & 1 &\ldots & 1-\lambda\end{vmatrix}&=\begin{vmatrix} 1-\lambda & \;\;1 & \ldots & \;\;1\\ \lambda &-\lambda & \ldots & \;\;0 \\ \vdots&&&\vdots \\ \lambda & \;\;0 &\ldots & -\lambda\end{vmatrix}\\&=\begin{vmatrix} n-\lambda & 1 & \ldots & 1\\ 0 &-\lambda & \ldots & 0 \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & -\lambda\end{vmatrix}\\&=(n-\lambda)(-\lambda)^{n-1}\end{aligned}

So, the eigenvalues are $\lambda=0$ (n-1-uple) and $\lambda=n$ (1-uple) if $n\geq 2$ . If $n=1$ we have only the eigenvalue $\lambda=1$ (1-uple).

I don't understand how you obtained $n - \lambda$ in the $a_{11}$ position.
• October 15th 2011, 03:41 PM
Deveno
Re: Eigenvalues
he started with 1-λ, and added (n-1) 1's. 1-λ + (n-1) = n-λ.
• October 15th 2011, 03:51 PM
dwsmith
Re: Eigenvalues
Quote:

Originally Posted by Deveno
he started with 1-λ, and added (n-1) 1's. 1-λ + (n-1) = n-λ.

How did he get the first column to be all zero after n-lambda? Why did only that entry change? How can we just add something to a_{11} and leave everything else alone?
• October 15th 2011, 04:00 PM
Deveno
Re: Eigenvalues
first he subtracted the first row, from every other row. this changed all the 1's in the first column (except for the top entry) to -λ.

it also changed all the 1's in every other column to 0, except along the diagonal, where 1-λ becomes -λ.

then he sucessively added (it doesn't matter if you do it one at a time, or all at once) columns 2 through n, to the first column.

neither the row operation, nor the column operation changes the determinant, since no row or column was multiplied.
• October 15th 2011, 04:03 PM
dwsmith
Re: Eigenvalues
Quote:

Originally Posted by Deveno
first he subtracted the first row, from every other row. this changed all the 1's in the first column (except for the top entry) to -λ.

it also changed all the 1's in every other column to 0, except along the diagonal, where 1-λ becomes -λ.

then he sucessively added (it doesn't matter if you do it one at a time, or all at once) columns 2 through n, to the first column.

neither the row operation, nor the column operation changes the determinant, since no row or column was multiplied.

Ok it was the adding of the columns I couldn't figure out.
• October 15th 2011, 04:08 PM
dwsmith
Re: Eigenvalues
Now the eigenvectors.

For $\lambda =0$, I obtained

$\left\{\begin{bmatrix}-1\\n\\0\\ \vdots\\0\end{bmatrix},\begin{bmatrix}-1\\0\\n\\ \vdots\\0\end{bmatrix},\cdots,\begin{bmatrix}-1\\0\\0\\ \vdots\\n\end{bmatrix}\right\}$

However, I am finding it hard to solve for lambda = n

Is it just

$\begin{bmatrix}1\\0\\0\\ \vdots\\0\end{bmatrix}$
• October 15th 2011, 04:40 PM
Deveno
Re: Eigenvalues
the eigenvalue n has a multiplicity of 1, so all we need to do is find ONE eigenvector. try (1,1,1,.....,1).

(how did i find this? note that (A-nI)(x1,x2,....,xn) = (0,0,....,0) implies that

x1+x2+....+xn = nxj for ALL j)
• October 15th 2011, 04:52 PM
dwsmith
Re: Eigenvalues
Quote:

Originally Posted by Deveno
the eigenvalue n has a multiplicity of 1, so all we need to do is find ONE eigenvector. try (1,1,1,.....,1).

(how did i find this? note that (A-nI)(x1,x2,....,xn) = (0,0,....,0) implies that

x1+x2+....+xn = nxj for ALL j)

Shouldn't we get this:

$\begin{vmatrix}0&1&\cdots & 1\\0&-n&\cdots &0\\ \vdots & 0& \ddots \\ 0&\cdots & &-n \end{vmatrix}$

$x_2=-x_3-x_4-\cdots -x_n$

$x_i=0 \ \ i=2,...,n$

$x_1$ is a free variable.

So shouldn't it just be the n dimensional $e_1$ vector?
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