Characteristic polynomial of a matrix

**jsridhar72** · August 11th 2010, 12:50 AM

I want to know what is the characteristic polynomial of a matrix.Please explain with simple example.What is the purpose of finding one?

**Ackbeet** · August 11th 2010, 02:38 AM

Let's take the matrix

$A=\begin{bmatrix}1&3\\5 &7\end{bmatrix}.$

The characteristic polynomial has to do with finding the eigenvalues of a matrix. The eigenvalues of a matrix are defined as follows:

The eigenvalues of a matrix $A$ are the numbers $\lambda$ such that $Ax=\lambda x$ for $x\not=0.$

Here $x$ is a vector of the appropriate size so that the matrix multiplication $Ax$ is defined. If you find such an $x$ and a $\lambda$ , then $x$ is called an eigenvector. $\lambda$ is, in general, a complex number.

The geometric interpretation of eigenvalues is that when you multiply an eigenvector $x$ by the matrix $A$ , it's the same thing as multiplying the eigenvector $x$ by a simple number. Multiplying by a simple number is basically stretching the eigenvector $x$ in some way. It doesn't change the eigenvector's direction. The whole eigenvalue problem, as we call it, has to do with finding a "shortcut" to matrix multiplication.

Why are we interested in the eigenvalue problem? Tons of reasons! You can use eigenvalues and eigenvectors (lump those two concepts together and call it the "eigensystem") to solve systems of ordinary differential equations, which in turn can model the behavior of an electrical circuit or a mass/spring system. This very problem shows up all over the place in Quantum Mechanics.

So, let's rearrange the eigenvalue problem quick, while nobody's looking:

$Ax-\lambda x=0.$

Now, I want to right-factor the $x$ out of there. But what's left had better all be of the same type: a matrix. So, I can insert the identity matrix in there to make it all the same:

$Ax-\lambda I x=0,$ where $I$ is the same size as $A$ . You'd agree I haven't changed anything, right? So now I get to factor (technically, right-factor: matrix multiplication is NOT, in general, commutative):

$(A-\lambda I)x=0.$

Now, remember: $x\not=0$ . This system looks like

$Bx=0,$ for $B=A-\lambda I.$

It's a fact from linear algebra that a homogeneous system can only have nonzero (nontrivial) solutions if the matrix $B$ is singular. Another way of saying that is that

$\det(B)=\det(A-\lambda I)=0.$ The equation

$\det(A-\lambda I)=0$ is the characteristic equation. The characteristic polynomial is the LHS of this equation.

So let's find this characteristic polynomial for our matrix $A$ above. Evidently, we need to form the matrix $A-\lambda I$ . Well, that gives us

$A-\lambda I= \begin{bmatrix}1&3\\5 &7\end{bmatrix}-\lambda \begin{bmatrix}1&0\\0 &1\end{bmatrix}= \begin{bmatrix}1&3\\5 &7\end{bmatrix}-\begin{bmatrix}\lambda&0\\0 &\lambda\end{bmatrix}= \begin{bmatrix}1-\lambda&3\\5 &7-\lambda\end{bmatrix}.$

Now that we've assembled that ingredient, we need to take its determinant and set it equal to zero:

$\det\begin{bmatrix}1-\lambda&3\\5 &7-\lambda\end{bmatrix}=(1-\lambda)(7-\lambda)-15=0.$

If that step mystifies you, you should review determinants. We can simplify this just a bit:

$(1-\lambda)(7-\lambda)-15=7-8\lambda+\lambda^{2}-15=\lambda^{2}-8\lambda-8=0.$

So the characteristic polynomial associated with our matrix $A$ is

$\lambda^{2}-8\lambda-8.$

To find the eigenvalues of our matrix $A$ , you'd set the characteristic polynomial equal to zero and solve:

$\lambda^{2}-8\lambda-8=0.$ The solutions are, of course,

$\displaystyle{\lambda=\frac{8\pm\sqrt{64-4(1)(-8)}}{2}=\frac{8\pm\sqrt{96}}{2}=\frac{8\pm 4\sqrt{6}}{2}=4\pm 2\sqrt{6}.}$

Does this all make sense?

**adam_leeds** · August 11th 2010, 03:49 AM

Is the characteristic polyomial the same as the characteristic equation?

**HallsofIvy** · August 11th 2010, 04:17 AM

No, because a "polynomial" is not an "equation"!

But if, say, the characteristic polynomial is $x^3- 3x^2+ x- 3$ then the characteristic equation is $x^3- 3x^2+ x- 3= 0$ .

**adam_leeds** · August 11th 2010, 04:25 AM

Originally Posted by HallsofIvy

No, because a "polynomial" is not an "equation"!

But if, say, the characteristic polynomial is $x^3- 3x^2+ x- 3$ then the characteristic equation is $x^3- 3x^2+ x- 3= 0$ .

Yeah i see, cheers.

**Ackbeet** · August 11th 2010, 04:33 AM

I've clarified the polynomial/equation distinction a bit more. I had it at the end of my post, but not in the middle. See the edited version.

**jsridhar72** · August 11th 2010, 10:15 PM

Originally Posted by Ackbeet

Let's take the matrix

$A=\begin{bmatrix}1&3\\5 &7\end{bmatrix}.$

The characteristic polynomial has to do with finding the eigenvalues of a matrix. The eigenvalues of a matrix are defined as follows:

The eigenvalues of a matrix $A$ are the numbers $\lambda$ such that $Ax=\lambda x$ for $x\not=0.$

Here $x$ is a vector of the appropriate size so that the matrix multiplication $Ax$ is defined. If you find such an $x$ and a $\lambda$ , then $x$ is called an eigenvector. $\lambda$ is, in general, a complex number.

The geometric interpretation of eigenvalues is that when you multiply an eigenvector $x$ by the matrix $A$ , it's the same thing as multiplying the eigenvector $x$ by a simple number. Multiplying by a simple number is basically stretching the eigenvector $x$ in some way. It doesn't change the eigenvector's direction. The whole eigenvalue problem, as we call it, has to do with finding a "shortcut" to matrix multiplication.

Why are we interested in the eigenvalue problem? Tons of reasons! You can use eigenvalues and eigenvectors (lump those two concepts together and call it the "eigensystem") to solve systems of ordinary differential equations, which in turn can model the behavior of an electrical circuit or a mass/spring system. This very problem shows up all over the place in Quantum Mechanics.

So, let's rearrange the eigenvalue problem quick, while nobody's looking:

$Ax-\lambda x=0.$

Now, I want to right-factor the $x$ out of there. But what's left had better all be of the same type: a matrix. So, I can insert the identity matrix in there to make it all the same:

$Ax-\lambda I x=0,$ where $I$ is the same size as $A$ . You'd agree I haven't changed anything, right? So now I get to factor (technically, right-factor: matrix multiplication is NOT, in general, commutative):

$(A-\lambda I)x=0.$

Now, remember: $x\not=0$ . This system looks like

$Bx=0,$ for $B=A-\lambda I.$

It's a fact from linear algebra that a homogeneous system can only have nonzero (nontrivial) solutions if the matrix $B$ is singular. Another way of saying that is that

$\det(B)=\det(A-\lambda I)=0.$ The equation

$\det(A-\lambda I)=0$ is the characteristic equation. The characteristic polynomial is the LHS of this equation.

So let's find this characteristic polynomial for our matrix $A$ above. Evidently, we need to form the matrix $A-\lambda I$ . Well, that gives us

$A-\lambda I= \begin{bmatrix}1&3\\5 &7\end{bmatrix}-\lambda \begin{bmatrix}1&0\\0 &1\end{bmatrix}= \begin{bmatrix}1&3\\5 &7\end{bmatrix}-\begin{bmatrix}\lambda&0\\0 &\lambda\end{bmatrix}= \begin{bmatrix}1-\lambda&3\\5 &7-\lambda\end{bmatrix}.$

Now that we've assembled that ingredient, we need to take its determinant and set it equal to zero:

$\det\begin{bmatrix}1-\lambda&3\\5 &7-\lambda\end{bmatrix}=(1-\lambda)(7-\lambda)-15=0.$

If that step mystifies you, you should review determinants. We can simplify this just a bit:

$(1-\lambda)(7-\lambda)-15=7-8\lambda+\lambda^{2}-15=\lambda^{2}-8\lambda-8=0.$

So the characteristic polynomial associated with our matrix $A$ is

$\lambda^{2}-8\lambda-8.$

To find the eigenvalues of our matrix $A$ , you'd set the characteristic polynomial equal to zero and solve:

$\lambda^{2}-8\lambda-8=0.$ The solutions are, of course,

$\displaystyle{\lambda=\frac{8\pm\sqrt{64-4(1)(-8)}}{2}=\frac{8\pm\sqrt{96}}{2}=\frac{8\pm 4\sqrt{6}}{2}=4\pm 2\sqrt{6}.}$

Does this all make sense?

Bit confusing ...Thank you very much...

**Ackbeet** · August 12th 2010, 01:56 AM

Bit confusing...

I'd be happy to break out portions that are confusing. Can you point them out for me, and I'll explain further?

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