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I have the following matrix, call it A: $\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{bmatrix}$
I need to find its eigenvalue using its characteristic polynomial. I already did this on smaller matrices which were comfortable to calculate.
But calculating an eigenvalue for this matrix seems too complex - Possibly there's a better method?

Edit: This needs to be done without an external help (e.g. Wolfram Alpha).

Marc van Leeuwen
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Dor
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    Are you sure you want to find the eigenvalues using the characteristic polynomial? – Git Gud Jan 10 '15 at 19:21
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    enter modern world: http://www.wolframalpha.com/input/?i=eigenvalues+%7B%7B1%2C1%2C1%2C1%7D%2C%7B1%2C1%2C1%2C1%7D%2C%7B1%2C1%2C1%2C1%7D%2C%7B1%2C1%2C1%2C1%7D%7D – janmarqz Jan 10 '15 at 19:24
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    but this is a symmetric rank one matrix so it has three eigenvalues zero and one equal to the trace of this matrix. – abel Jan 10 '15 at 19:25
  • more: http://www.wolframalpha.com/input/?i=characteristic+polynomial+%7B%7B1%2C1%2C1%2C1%7D%2C%7B1%2C1%2C1%2C1%7D%2C%7B1%2C1%2C1%2C1%7D%2C%7B1%2C1%2C1%2C1%7D%7D – janmarqz Jan 10 '15 at 19:26

2 Answers2

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An easier method: note that this is a symmetric rank one matrix. In fact, let $x = (1,1,1,1)^T \in \Bbb R^4$. Note that $A = xx^T$.

What does the kernel of $A$ look like? What does this tell you about its eigenvalues?

Alternatively: note that $Ay = x(x^Ty) = \langle y,x \rangle x$. Perhaps you can use this to deduce the eigenvectors of $A$ directly.

Alternatively: let $u_1 = \frac 12 x$, which has length $1$. Extend $u_1$ to an orthonormal basis $u_1,\dots,u_4$. Let $U$ be the matrix whose columns are these vectors. Note that $$ A = U\pmatrix{4\\&0\\&&0\\&&&0}U^T $$

Ben Grossmann
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  • The kernel of A has dimension 3... It doesn't tell me anything about its eigenvalue :/ – Dor Jan 10 '15 at 19:40
  • @Dor yes it does. It tells you that $A$ has the eigenvalue $0$ with geometric multiplicity $3$. What else could we deduce, using perhaps the fact that $A$ is symmetric? – Ben Grossmann Jan 10 '15 at 19:40
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Set $x = (1, 1,1,1)^\top$, so that $A= xx^\top$. Then $$ \det(A - \lambda I) = (-\lambda)^4 \det(I - xx^\top/\lambda).$$ By Sylvester's Determinant Theorem, this equals $$(-\lambda)^4 \det(I - x^\top x/\lambda) = (-\lambda)^4 (1-4/\lambda) = \lambda^3(\lambda-4). $$ From there, you're home free.

Avi Steiner
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