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Zinaida [17]
3 years ago
15

Let the matrix below act on C². Find the eigenvalues and a basis for each eigenspace in C².

Mathematics
1 answer:
forsale [732]3 years ago
4 0

Hello, let's note A the matrix, we need to find \lambda such that A\lambda=\lambda I, where I is the identity matrix, so the determinant is 0, giving us the characteristic equation as

\left|\begin{array}{cc}1-\lambda&3\\-3&1-\lambda\end{array}\right|\\\\=(1-\lambda)^2+9\\\\=\lambda^2-2\lambda+10\\\\=0

We just need to solve this equation using the discriminant.

\Delta=b^2-4ac=2^2-40=-36=(6i)^2

And then the eigenvalues are.

\lambda_1=\dfrac{2-6i}{2}=\boxed{1-3i}\\\\\lambda_2=\boxed{1+3i}

To find the basis, we have to solve the system of equations.

A\lambda_1-\lambda_1 I=\left[\begin{array}{cc}3i&3\\-3&3i\end{array}\right] \\\\=3\left[\begin{array}{cc}i&1\\-1&i\end{array}\right] \\\\\text{For a vector (a,b), we need to find a and b such that.}\\\\\begin{cases}ai+b=0\\-a+bi=0\end{cases}\\\\\text{(1,-i) is a base of this space, as i-i=0 and -1-}i^2\text{=-1+1=0.}

A\lambda_2-\lambda_2 I=\left[\begin{array}{cc}-3i&3\\-3&-3i\end{array}\right] \\\\=3\left[\begin{array}{cc}-i&1\\-1&-i\end{array}\right]\\\\\text{For a vector (a,b), we need to find a and b such that.}\\\\\begin{cases}-ai+b=0\\-a-bi=0\end{cases}\\\\\text{(1,i) is a base of this space as -i+i=0 and -1-i*i=0.}

Thank you

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Answer:

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You didn't include the costs of doing repairs at both computer repair firms. I fixed that for you.

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Step-by-step explanation:

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