> For the complete documentation index, see [llms.txt](https://veriny.gitbook.io/berkeley/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://veriny.gitbook.io/berkeley/math-270-linear-algebra/chapter-5-eigenvectors-and-eigenvalues.md). # Eigenvectors and Eigenvalues In this section, we will discuss eigenvectors, eigenvalues (**λ**) and eigenspaces. λ is an eigenvalue of matrix A given the following condition. $$ Ax=λx $$ Here it is rewritten in a more useful way. $$ (A-λI)x=0 $$ Where A is an m x n matrix, x is a n x 1 matrix, and λ is a scalar. Similarly, eigenvectors corresponding to all the λ of A are all the vectors x such that the above condition is bet. Together with the 0 vector, the eigenvectors are a subspace of R^n, referred to as the eigenspace. {% hint style="info" %} Eigenspaces must have at least one nontrivial basis vector. 0 alone cannot be an eigenspace. {% endhint %} Here are brief summaries regarding how to find all three. ## Finding Eigenvectors/Eigenspaces Given the following matrix $$ (A-λI) $$ The **null space** of this matrix corresponds to the eigenspace of A corresponding to the λ used. That is, augment a column of zeroes and row reduce it. Then you can find the null space. {% hint style="info" %} Note that there are commonly more than one eigenvalues. {% endhint %} It follows that each vector in the eigenspace of A corresponds to an eigenvector. ## Finding Eigenvalues We know that the matrix discussed in the previous section, (A - λI), has a null space, meaning that (A - λI)x = 0 has some non-trivial solution. From this, we know that (A - λI) is not invertible (by the Invertible Matrix Theorem, see previous section). Therefore, $$ det(A-λI)=0 $$ When simplifying this, you will find that you get a polynomial with λ as the variable. Simply find the solutions to the polynomial and you will have your eigenvalues. {% hint style="info" %} If A is a triangular matrix, the eigenvalues are the entries on the main diagonal. {% endhint %} {% hint style="warning" %} If 0 is an eigenvalue of A, A is not invertible. {% endhint %} {% hint style="info" %} λ^-1 is an eigenvalue of A^-1 {% endhint %} ## Diagonal Matrices and Similar Matrices Somewhat obviously, a diagonal matrix is a matrix in which all non-diagonal entires are zeroes. A matrix A is diagonalizable if A is similar to a diagonal matrix. A is said to be similar to a diagonal matrix D if there is an matrix P such that the following holds true. $$ D=P^{-1}AP $$ Another form of this equation is — $$ PDP^{-1}=A $$ {% hint style="danger" %} If an n x n matrix does not have n distinct values, it is not diagonalizable. Recognizing this early can save a lot of time. {% endhint %} ## Important Theorems to Remember Following are some important theorems to remember regarding eigenvectors, eigenvalues, and diagonal matrices. * The dimension of the eigenspace of A is equal to the multiplicity (that is, the highest degree) of λ. * A is diagonalizable if the sum of the dimensions of all of the eigenspaces of A equals n. * If A is diagonalizable and $$B\_k$$ is a basis for the eigenspace corresponding to λ\_k for each k, then the total collection of vectors in the sets B\_1, B\_2, B\_k forms an eigenvector basis for R^n. * The following holds true: $$ A^k=PD^kP^{-1} $$ * In $$A=PDP^{-1},$$ the columns of P are the n Linearly independent eigenvectors of A as well as the diagonal entries of D being the corresponding eigenvalues of A. * A square matrix is diagonalizable if $$A$$ has exactly $$n$$ linearly independent eigenvectors. That is, the sum of all the bases for all the eigenspaces equals $$n$$ , * If 0 is an eigenvalue of $$A$$ , it is **not** invertible. * Given A is a 2 x 2 matrix with complex eigenvalues, then given $$\lambda=a-bi$$ , in the formula $$A=PCP^{-1}$$ $$ P=\[Rv\ \ Imv]\\ C=\begin{bmatrix} a&-b\\ b\&a \end{bmatrix} $$ * When finding the eigenvectors for a complex eigenvalue, either the first or second row of $$\[A-\lambda\_1I | 0]$$ may be used. * Eigenvectors that correspond to distinct eigenvalues are linearly independent. * Eigenvectors for complex conjugate eigenvalues are also complex conjugates. * The matrix A is diagonalizable if the dimension of the eigenspace of each $$\lambda\_k$$ is equal to the multiplicity of $$\lambda\_k$$ ## A Return to Polynomial Subspaces A lot of the "new" material that is introduced here is actually fairly intuitive. If **T** is a linear transformation from v to w, in accordance with the below diagram — ![Credit: Behrooz Shahrvini](/files/-MC5H32p2ZnxQ6L7evVp) — then the following holds true. $$ \[T(x)]\_c=m \*\[x]\_b $$ where $$ \[\[T(b\_1)\_c],\[T(b\_2)\_c]...\[T(b\_n)\_c]] $$ This section is probably best shown using examples, so we'll use this one. Just make sure to keep track of all the vocabulary. Find the **image** of p(t) = 3-2t+t^2 given T(p(t)) = p(t)+2t^2p(t) This, of course, means that we plug p(t) into the transformation and see what we get. Show that T is a **linear transformation.** For this, we can simply ensure that it is closed under scalar multiplication and addition. $$ T(P\_1(t)+P\_2(t))=T(P\_1(t))+T(P\_2(t))\\ T(cP(t)) = cT(P(t)) $$ Plug and chug, pretty much. Find M, the matrix for T relative to the bases B and C. Probably should have mentioned this earlier, but given — $$ B={1, t, t^2}\\ C={1.t.t^2.t^3.t^4} $$ From here, it's a simple matter of plugging each value into T(p(t)) for p(t). For simplicity's sake, only the t^2 value is demonstrated for B. $$ T(t^2)=t^2+2t^2(t^2) $$ We would do that for each thing in each basis and turn the solutions into vectors, which can then be joined into a matrix that represents M. ## B-matrices The B-matrix for a linear transformation given A and a set of vectors B can be found with the following method. Given: $$ B={b\_1, b\_2}\\ b\_1= \begin{bmatrix} 2\\ 1 \end{bmatrix} b\_2= \begin{bmatrix} 1\\ 1 \end{bmatrix}\\ A=\begin{bmatrix} -2&-1\\ 3&1 \end{bmatrix} $$ P is given as follows $$ P=\begin{bmatrix} 2&1\\ 1&1 \end{bmatrix} $$ And the B-matrix can be found with the following formula. $$ \[T]\_b=P^{-1}AP $$ {% hint style="success" %} As we learned earlier, \[T]\_b is similar to A by the definition of a similar matrix. {% endhint %} Because of the above property, we can actually find a basis B for R\_2 as well as \[T]\_b given *only* A. We simply assume that \[T]\_b is diagonal, meaning that we can find the eigenvalues of A and simply arrange them in a diagonal matrix. $$ \[T]\_b=\begin{bmatrix} λ\_1&0\\ 0&λ\_2 \end{bmatrix} $$ {% hint style="danger" %} Note: when reorganizing matrix equations by moving matrices to one side or the other, if a matrix is right multiplied originally it must st right multiplied when moved to the other side, as well as vice versa. {% endhint %} ## Angle of Rotation The angle of rotation $$\Phi$$ can be found with the equation $$\arctan{\frac{b}{a}}$$ . If a = 0, you can still find the angle if you use $$\arcsin{b}$$ although make sure you divide out the scaling factor or you may run into some trouble. --- # Agent Instructions This documentation is published with GitBook. GitBook is the documentation platform designed so that both humans and AI agents can read, navigate, and reason over technical content effectively. 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