# Finding the eigenvectors for T()

In summary: Therefore, (-2, -2) is an eigenvector with eigenvalue 2. Similarly, T(1, -1) = (0, 0) = 0*(1, -1), so (1, -1) is an eigenvector with eigenvalue 0. And T(1, 2) = (3, 3) is not equal to any scalar multiple of (1, 2), so (1, 2) is not an eigenvector. Therefore, the answer is C.

## Homework Statement

Which of the following is not an eigenvector for $$T \left( \left[ {\begin{array}{cc} x \\ y \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} x + y \\ x+ y \\ \end{array} } \right]$$ ?

A) v = [-2 -2]T
B) v = [1 -1]T
C) v = [1 2]T
D) All are eigenvectors

## Homework Equations

Ax = $$\lambda$$x

## The Attempt at a Solution

My problem is that all eigenvectors I've computed have come from 2x2 matrices. My best guess on starting is
$$T \left( \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} -4 \\ -4 \\ \end{array} } \right] \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right] =$$

but this obviously doesn't work because of the size. How do I find an eigenvalue of a 2x1 matrix? Is it possible? Am I even looking at this correctly?

Edit: I think I've figured it out. First, I constructed the standard matrix for T and got $$\left[ {\begin{array}{cc} 1&1 \\ 1&1 \\ \end{array} } \right]$$
Then I used Ax = $$\lambda$$x with the vectors given to find the eigenvalues. Letter C didn't have an eigenvalue, so that is the answer.
$$T \left( \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} 1&1 \\ 1&1 \\ \end{array} } \right] \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right] = \left[ {\begin{array}{cc} -4 \\ -4 \\ \end{array} } \right] = 2 \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right]$$
$$T \left( \left[ {\begin{array}{cc} 1 \\ -1 \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} 1&1 \\ 1&1 \\ \end{array} } \right] \left[ {\begin{array}{cc} 1 \\ -1 \\ \end{array} } \right] = \left[ {\begin{array}{cc} 0 \\ 0 \\ \end{array} } \right] = 0 \left[ {\begin{array}{cc} 1 \\ -1 \\ \end{array} } \right]$$
$$T \left( \left[ {\begin{array}{cc} 1 \\ 2 \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} 1&1 \\ 1&1 \\ \end{array} } \right] \left[ {\begin{array}{cc} 1 \\ 2 \\ \end{array} } \right] = \left[ {\begin{array}{cc} 3 \\ 3 \\ \end{array} } \right] = ?$$

Last edited:

## Homework Statement

Which of the following is not an eigenvector for $$T \left( \left[ {\begin{array}{cc} x \\ y \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} x + y \\ x+ y \\ \end{array} } \right]$$ ?

A) v = [-2 -2]T
B) v = [1 -1]T
C) v = [1 2]T
D) All are eigenvectors

## Homework Equations

Ax = $$\lambda$$x

## The Attempt at a Solution

My problem is that all eigenvectors I've computed have come from 2x2 matrices. My best guess on starting is
$$T \left( \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} -4 \\ -4 \\ \end{array} } \right] \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right] =$$

but this obviously doesn't work because of the size. How do I find an eigenvalue of a 2x1 matrix? Is it possible? Am I even looking at this correctly?

Edit: I think I've figured it out. First, I constructed the standard matrix for T and got $$\left[ {\begin{array}{cc} 1&1 \\ 1&1 \\ \end{array} } \right]$$
Then I used Ax = $$\lambda$$x with the vectors given to find the eigenvalues. Letter C didn't have an eigenvalue, so that is the answer.
$$T \left( \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} 1&1 \\ 1&1 \\ \end{array} } \right] \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right] = \left[ {\begin{array}{cc} -4 \\ -4 \\ \end{array} } \right] = 2 \left[ {\begin{array}{cc} -2 \\ -2 \\ \end{array} } \right]$$
$$T \left( \left[ {\begin{array}{cc} 1 \\ -1 \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} 1&1 \\ 1&1 \\ \end{array} } \right] \left[ {\begin{array}{cc} 1 \\ -1 \\ \end{array} } \right] = \left[ {\begin{array}{cc} 0 \\ 0 \\ \end{array} } \right] = 0 \left[ {\begin{array}{cc} 1 \\ -1 \\ \end{array} } \right]$$
$$T \left( \left[ {\begin{array}{cc} 1 \\ 2 \\ \end{array} } \right] \right) = \left[ {\begin{array}{cc} 1&1 \\ 1&1 \\ \end{array} } \right] \left[ {\begin{array}{cc} 1 \\ 2 \\ \end{array} } \right] = \left[ {\begin{array}{cc} 3 \\ 3 \\ \end{array} } \right] = ?$$

Your answer is correct, but you have made extra work for yourself in this problem. You didn't need to find a matrix representation for this transformation. All you needed to do was determine whether T(x) is equal to a scalar multiple of x.
For example, T(-2, -2) = (-4, -4) = 2*(-2, -2).

## 1. What are eigenvectors and why are they important in science?

Eigenvectors are special vectors that represent the directions in which a linear transformation (such as T()) acts by simply scaling the vector, without changing its direction. They are important in science because they allow us to understand the behavior and patterns of complex systems, and can be used to solve a wide range of problems in fields such as physics, engineering, and data analysis.

## 2. How do you find the eigenvectors for T()?

To find the eigenvectors for T(), we first need to find the eigenvalues, which are the scalars that represent the amount by which the eigenvectors are scaled when acted upon by T(). This can be done by solving the characteristic equation det(T() - λI) = 0, where I is the identity matrix. Once the eigenvalues are found, we can then solve for the corresponding eigenvectors by plugging in each eigenvalue into the equation T()v = λv and solving for v.

## 3. Can a matrix have more than one eigenvector?

Yes, a matrix can have multiple eigenvectors for a single eigenvalue. In fact, the number of linearly independent eigenvectors for a given eigenvalue is called the geometric multiplicity, and it can be less than, equal to, or greater than the algebraic multiplicity (the number of times that eigenvalue appears as a root of the characteristic equation).

## 4. What is the relationship between eigenvectors and eigenvalues?

Eigenvectors and eigenvalues are closely related, as each eigenvector corresponds to a specific eigenvalue. The eigenvalue represents the amount by which the eigenvector is scaled when acted upon by the linear transformation. Additionally, the eigenvectors corresponding to distinct eigenvalues are orthogonal (perpendicular) to each other, which can be useful in certain applications.

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

Eigenvectors are commonly used in data analysis techniques such as Principal Component Analysis (PCA) and Singular Value Decomposition (SVD). In these methods, the eigenvectors are used to transform the data into a new coordinate system, where the first few eigenvectors capture the most variation in the data. This allows for dimensionality reduction and identification of important patterns and trends in the data.

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