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Let, $A=\begin{bmatrix} 4&0&1&0\\1&1&1&0\\0&1&1&0 \\0&0&0&4 \end{bmatrix}$. Knowing that $4$ is one of its eigenvalues, find the characteristic polynomial of $A$.

Well if $4$ is an eigenvalues of $A$, one should have $|A-4I_{4}|=0$ . And so,

$\begin{vmatrix} 0&0&1&0\\1&-3&1&0\\0&1&-3&0 \\0&0&0&0 \end{vmatrix}=0$

It's clear that the previous equation is true (the determinant of $(A-4I_{4})=0$). Now that the factor $(\lambda-4)$ was pull out, one gets a new matrix by removing the null row and null column.

$A'=\begin{bmatrix} 0&0&1\\1&-3&1\\0&1&-3&\end{bmatrix}$

The characteristic polynomial of $A'$ will be a $3^{th}$ degree polynomial, which product with $(\lambda-4)$ equals to a $4^{th}$ degree polynomial.

Now, in order of finding the characteristic polynomial of $A'$ one must to solve the characteristic equation:

$\begin{vmatrix} -\lambda&0&1\\1&-3-\lambda&1\\0&1&-3-\lambda&\end{vmatrix}=0$

My doubt is on finding this determinant. I already tryed Laplace's transformations in order to make null row or a column, but I couldn't do it.

Can you give me a clue? Thanks.

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4  
Just use the formula for $\det$. –  copper.hat Mar 26 at 18:12

4 Answers 4

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...and how to find the charateristic polynomial of the original matrix, $A$ $$\begin{align}\mathrm{det}(A - \lambda \mathrm{I}) &= 0 \tag{1}\\ \begin{vmatrix}(4-\lambda)&0&1&0 \\ 1&(1-\lambda)&1&0\\ 0&1&(1-\lambda)&0\\ 0&0&0&(4-\lambda) \end{vmatrix} &= 0 \tag{2}\\ (4-\lambda)\begin{vmatrix}(4-\lambda)&0&1\\ 1&(1-\lambda)&1\\ 0&1&(1-\lambda) \tag{3}\\ \end{vmatrix} &= 0 \\ (4-\lambda) \left[(4-\lambda) \left[(1-\lambda)^2 -1 \right] + 1\right] &= 0\tag{4} \\ (4-\lambda) \left[(4-\lambda)(\lambda^2 -2\lambda) + 1\right] &= 0\tag{5} \\ (4-\lambda) \left[(4\lambda^2 -8\lambda -\lambda^3+2\lambda^2) + 1\right] &= 0 \tag{6}\\ (4-\lambda)(-\lambda^3 + 6\lambda^2 - 8\lambda +1 ) &= 0 \tag{7}\\ (\lambda -4)(\lambda^3 - 6\lambda^2 + 8\lambda - 1)&= 0\tag{8}\\ \end{align}$$

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$A$ does not have repeated eigenvalues. Going from $4^{\mathrm{th}}$ to $5^{\mathrm{th}}$ equation is incorrect. –  John Habert Mar 26 at 21:48
    
@JohnHabert Thank You! - will fix that. –  Brad S. Mar 26 at 21:59
    
@BradS. so my assumption that the $P_{A}(\lambda)=(\lambda-\alpha) \cdot P_{A'}(\lambda)$ is always false? Or in this case is false? Thanks. –  João Mar 26 at 22:20
    
@João please check my work. I have made many algebra errors. Hopefully, they have all been corrected now. (sigh). I didn't sleep well last night. I think I'll go home now. Hope some of this helped more than it confused. –  Brad S. Mar 26 at 22:34
1  
I think you are off by just a sign. The final polynomial is what you have with all signs flipped. @João You should always find characteristic polynomial from the original. –  John Habert Mar 26 at 22:37

You can expand the determinant along the first row (see http://en.wikipedia.org/wiki/Determinant#3.C2.A0.C3.97.C2.A03_matrices and http://en.wikipedia.org/wiki/Laplace_expansion) and get $\det A' = -\lambda ((-3 -\lambda)^2 - 1) - 0 + 1\cdot1$

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Working across the top row.... $$\begin{align}\begin{vmatrix} -\lambda&0&1\\1&-3-\lambda&1\\0&1&-3-\lambda&\end{vmatrix} &= -\lambda\left[(-3 -\lambda)(-3 -\lambda) -1\right] -0[\,] + 1\left[1-0 \right]=0 \\ &= -\lambda\left[9 + 6\lambda +\lambda^2 -1 \right] + 1= 0 \\ &= -\lambda^3 -6\lambda^2 -8\lambda +1 = 0\\ &= \lambda^3 +6\lambda^2 +8\lambda -1 = 0 \end{align}$$

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1  
This method is correct for obtaining the characteristic polynomial of $A'$, but not for finding the remaining eigenvalues or characteristic polynomial of $A$. See my answer below. –  Robert Lewis Mar 26 at 20:08
    
@RobertLewis but, the OP only asked how to find the determinate shown here. –  Brad S. Mar 26 at 20:19
    
Agreed! I tried to make this clear. –  Robert Lewis Mar 26 at 20:25

As rschwieb pointed out in his answer, which has mysteriously disappeared since my original posting of this answser, the method proposed by the OP in the text of the question is erroneous; indeed, the first answer proposed by Brad S. is also problematic in terms of finding the characteristic polynomial of $A$, though his calculations are apparently correct for the characteristic polynomial of $A'$, which is the issue he is evidently addressing.

To explicitly use the fact that $4$ is an eigenvalue of $A$, observe that the $4$ in the lower-right hand corner of $A$ corresponds to the eigenvector $\mathbf e_4 = (0, 0, 0, 1)^T$, and that the upper left $3 \times 3$ block of $A$, hence of $A - \lambda$, leaves the subspace normal to $\mathbf e_4$ invariant. Thus the eigenvalues of $A$ restricted to that invariant subspace will simply be those of $A$ exclusive of the eigenvalue $4$ associated with $\mathbf e_4$; the characteristic polynomial of this restriction of $A$ is in fact

$\det(\begin{bmatrix} 4 - \lambda &0&1\\1&1 - \lambda &1\\0&1&1 - \lambda \end{bmatrix}) = (4 - \lambda)(\lambda^2 - 2\lambda) + 1 = -\lambda^3 + 6\lambda^2 - 8\lambda + 1, \tag{0}$

as is shown in detail below. Multiplying (0) by the factor $4 - \lambda$ yields $p_A(\lambda) = (4 - \lambda)(-\lambda^3 + 6\lambda^2 - 8\lambda + 1)$ for the characteristic polynomial of $A$.

It appears to me that the error in the OP's method lies in computing the eigenvalues of $A - 4I$ on the subspace spanned by $\mathbf e_1 = (1, 0, 0)^T$, $\mathbf e_2 = (0, 1, 0)^T$, $\mathbf e_3 = (0, 0, 1)^T$, instead those of the restriction of $A$ itself. Indeed, an inspection of rschwieb's tables, when they existed, showed that the eigenvalues of $A'$ may be had by subtracting $4$ from those of $A$, exclusive of the eigenvalue $\lambda = 4$ of $A$. This agrees with the basic fact that $B\mathbf v = \mu \mathbf v \Leftrightarrow (B + \alpha I)v = (\mu + \alpha)\mathbf v$, i.e., eigenvalues simply shift by $\alpha$ if $\alpha I$ is added to a matrix $B$.

Having said these things, here's how I would handle this one:

the given matrix is

$A=\begin{bmatrix} 4&0&1&0\\1&1&1&0\\0&1&1&0 \\0&0&0&4 \end{bmatrix}, \tag{1}$

so

$A - \lambda I = \begin{bmatrix} 4 - \lambda &0&1&0\\1&1 - \lambda &1&0\\0&1&1 - \lambda&0 \\0&0&0&4 - \lambda \end{bmatrix}, \tag{2}$

so the characteristic polynomial $p_A(\lambda)$ is

$p_A(\lambda) = \det (A - \lambda I) = \det (\begin{bmatrix} 4 - \lambda &0&1&0\\1&1 - \lambda &1&0\\0&1&1 - \lambda&0 \\0&0&0&4 - \lambda \end{bmatrix}); \tag{3}$

since the last row and last column are $0$, save for the $4, 4$ entry, expansion in minors along the last row or column yields

$p_A(\lambda) = (4 - \lambda) \det(\begin{bmatrix} 4 - \lambda &0&1\\1&1 - \lambda &1\\0&1&1 - \lambda \end{bmatrix}) \tag{4}$

and we have, by any of a number of standard methods/formulas for the computation of $3 \times 3$ determinants

$\det(\begin{bmatrix} 4 - \lambda &0&1\\1&1 - \lambda &1\\0&1&1 - \lambda \end{bmatrix}) = (4 - \lambda)(1 - \lambda)^2 + 1 - (4 - \lambda)$ $= (4 - \lambda)((1 - \lambda)^2 - 1) + 1 = (4 - \lambda)(\lambda^2 - 2\lambda) + 1$ $= -\lambda^3 + 6\lambda^2 - 8\lambda + 1, \tag{5}$

so that

$p_A(\lambda) = (4 -\lambda)(-\lambda^3 + 6\lambda^2 - 8\lambda + 1) = \lambda^4 - 10\lambda^3 + 32\lambda^2 - 33\lambda + 4. \tag{6}$

This method tacitly exploits the fact that $4$ is an eigenvalue of $A$ with eigenvector $\mathbf e_4$ when it invokes the expansion by minors; the fact that $A \mathbf e_4 = 4 \mathbf e_4$ is related to the zeroes of the last row and column of $A$; but about this I will say no more at present. My day job beckons.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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@John Habert: thanks, I'll check that out! –  Robert Lewis Mar 26 at 20:11

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