# How to find eigenvalues/eigenvectors

• mind0nmath
In summary, to find the eigenvalues and eigenvectors for the linear operator T defined as T(w,z) = (z,w), you can represent T as a matrix and apply it to each basis vector to form the columns of the matrix. Alternatively, you can use the definition of eigenvalues and find values of lambda that satisfy the equations for non-zero solutions. Using these methods, we can determine that the matrix representing T in the standard basis has all 1s, and its eigenvalues are 0 and 2.
mind0nmath
How do i find the eigenvalues and eigenvectors for the linear operator T defined as
T(w,z) = (z,w)??

mind0nmath said:
How do i find the eigenvalues and eigenvectors for the linear operator T defined as
T(w,z) = (z,w)??

I'd start by writing T as a matrix.

how about for something like: T(x_1,x_2,...,x_n) = (x_1+x_2+...+x_n, x_1+x_2+...+x_n, ..., x_1+x_2+...+x_n). The matrix with respect to standard basis would have 1's everywhere? any clues to finding the eigenvalues/vectors?

Try Matlab command >>[V,D] = eig(ones(n))

? T operates on a pair of numbers and gives a pair of numbers as the result. Written as a matrix, it would be 2 by 2 matrix- certainly not as complicated as you have! You are not still referring to the first problem are you?

By definition, T(w,z)= (z, w) so T(1, 0)= (0, 1) and T(0, 1)= (1, 0).
$$\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\left(\begin{array}{c} 1 \\ 0\end{array}\right)= \left(\begin{array}{c} 0 & 1\end{array}\right)$$
What are a and c? Do the same with (0, 1) being taken to (1, 0) to determine b and d.
A good way of determining the matrix representing a linear operator in a given basis is to apply it to each of the basis vectors in turn. The result will be a column of the matrix.

Of course, you don't have to write it as a matrix to find eigenvalues- in fact, I think too many students get the idea that Linear Algebra is only about matrices. Saying that $\lambda$ is an eigenvalue for linear transformation T means that there exist some (x, y), not both 0, such that $T(x,y)= \lambda(x, y)= (\lambda x, \lambda y)$. Since T(x,y)= (y, x), that says that $(y, x)= (\lambda x, \lambda y)$ so you have two equations: $y= \lambda x$ and $x= \lambda y$. Obviously, x= y= 0 would satisfy those equations for any $\lambda$. For what values of $\lambda$ would that have non-zero solutions? If you replace the "x" in the first equation by $\lambda y$ from the second equation, you have $y= \lambda(\lambda y)= \lambda^2 y$. If y is not 0, you can divide both sides by y to get $\lambda^2= 1$.

mind0nmath said:
how about for something like: T(x_1,x_2,...,x_n) = (x_1+x_2+...+x_n, x_1+x_2+...+x_n, ..., x_1+x_2+...+x_n). The matrix with respect to standard basis would have 1's everywhere? any clues to finding the eigenvalues/vectors?
One of the things you should have learned long ago is that you approach problems like this by looking at simple cases: if n= 2, this says T(x,y)= (x+ y, x+ y). In particular, T(1, 0)= (1, 1) and T(0,1)= (1, 1). Yes, the columns of the matrix representing this linear operator in the standard basis are all 1s. The matrix representing this linear operator in the standard basis consists of all 1s.

Okay, what are the eigenvalues of
$$\left(\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right)$$?
(Hint: if a matrix has two rows (or two columns) the same it has determinant 0. And if it has deteminant 0, it has 0 as an eigenvalue.)

The eigenvalues must satisfy
$$\left|\begin{array}{cc} 1-\lambda & 1 \\ 1 & 1- \lambda\end{array}\right|= 0$$

What equation does that give you? What are the eigenvalues?

it seems obvious that (1,1) goes to (1,1), and (1,-1) goes to? so the eigenvalues are...

## 1. What are eigenvalues and eigenvectors?

Eigenvalues and eigenvectors are concepts in linear algebra that are used to represent the characteristics of a matrix. Eigenvalues are scalar values that represent the amount by which an eigenvector is scaled when multiplied by a matrix. Eigenvectors are non-zero vectors that, when multiplied by a matrix, result in a scalar multiple of the original vector.

## 2. Why are eigenvalues and eigenvectors important?

Eigenvalues and eigenvectors are important because they are used to solve many problems in linear algebra, such as finding the roots of a polynomial or solving systems of differential equations. They also have applications in fields such as physics, engineering, and computer science.

## 3. How do I find eigenvalues and eigenvectors?

To find eigenvalues and eigenvectors, you first need to find the characteristic polynomial of the matrix. This is done by subtracting the variable λ from the diagonal elements of the matrix and finding the determinant. The eigenvalues are then the values of λ that make the determinant equal to 0. Once you have the eigenvalues, you can find the eigenvectors by solving a system of equations using the eigenvalues.

## 4. Can all matrices have eigenvalues and eigenvectors?

No, not all matrices have eigenvalues and eigenvectors. Only square matrices (matrices with the same number of rows and columns) have eigenvalues and eigenvectors. Additionally, not all square matrices have distinct eigenvalues, which means they may have repeated or complex eigenvalues.

## 5. How are eigenvalues and eigenvectors used in data analysis?

In data analysis, eigenvalues and eigenvectors are used in techniques such as principal component analysis (PCA) and singular value decomposition (SVD). These methods use the eigenvalues and eigenvectors of a matrix to reduce the dimensionality of data and identify patterns and relationships between variables.

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