MAT223: 15&16_Eigenvectors and Diagonalization

Eigenvalues, Eigenvectors and Diagonalization §

Eigenvalues and Eigenvectors §

• Eigenvector - Let $X$ be a linear transformation or a matrix. An eigenvector for $X$ is a non-zero vector that doesn’t change the directions when $X$ is applied. That is $\vec{v}\ne \vec{0}$ is an eigenvector for $X$:
  $$X\vec{v}=\lambda \vec{v}$$
• Eigenvalue - is a scaler $\lambda$, of $X$ corresponding to the eigenvector $\vec{v}$

Finding Eigenvectors §

• Let $M$ be a square matrix. The vector $\vec{v}\ne \vec{0}$ is an eigenvector for $M$ if and only if there exist a scaler $\lambda$ so that
  $$M\vec{v}=\lambda \vec{v}\Rightarrow M\vec{v}-\lambda \vec{v}=(M-\lambda I)\vec{v}=\vec{0}$$
  Now have that $\vec{v}\ne \vec{0}$ is an eigenvector for $M$ if and only if
  $$E_{\lambda }\vec{v}=(M-\lambda I)\vec{v}=M\vec{v}-\lambda \vec{v}=\vec{0}\_,$$

Characteristics Polynomial §

• Characteristic Polynomial - for a matrix $A$, the characteristic polynomial of $A$ is
  $$\text{char}(A)=\text{dec}(A-\lambda I)$$
• For an $n\times n$ matrix $A$, $char(A)$ has some nice properties
  • $char(A)$ is a polynomial
  • $char(A)$ has degree $n$
  • The coefficient of the ${\lambda }^n$ term in $char(A)$ is $\pm 1$; $+1$ if $n$ is even and $-1$ if $n$ is odd
  • $char(A)$ evaluated at $\lambda =0$ is $det(A)$
  • The roots of $char(A)$ are precisely the eigenvalues of $A$

Transformations without eigenvectors §

• If $A$ is a square matrix, then $A$ always has an eigenvalue provided complex eigenvalues are permitted.

Diagonalization §

• Diagonalization - a matrix is diagonalizable if it is similar to a diagonal matrix
• A linear transformation $\mathcal{T}:{\mathbb{R}}^n\to {\mathbb{R}}^n$ can be represented by a diagonal matrix if and only if there exists a basis for ${\mathbb{R}}^n$ consisting of eigenvectors for $\mathcal{T}$. If $\mathcal{B}$ is such a basis, then $[\mathcal{T}{]}_{\mathcal{B}}$ is a diagonal matrix

Non-diagonalizable matrices §

• Not every matrices are diagonalizable, but can check if an $n\times n$ matrix is diagonalizable by determining whiter there is a basis of eigenvectors for ${\mathbb{R}}^n$
  • If a matrix $A$ does not have real roots for $char(A)$, then it has no real eigenvalues, so not diagonalizable
  • If a matrix $J$ has eigenvectors that is not contain the its domain, then $J$ is not diagonalizable

Geometric and Algebraic Multiplicities §

• Eigenspace - Let $A$ be a $n\times n$ matrix with eigenvalues ${\lambda }_1,\dots ,{\lambda }_m$. The eigenspace of $A$ corresponding to he eigenvalues of ${\lambda }_i$ is the null space of $A-{\lambda }_{i}I$. That si the space spanned by all eigenvectors that have the eigenvalues ${\lambda }_i$

  • The geometric multiplicity of eigenvalues ${\lambda }_i$ is the dimension of the corresponding eigenspace.

  • The algebraic multiplicity of ${\lambda }_i$ is the number of times ${\lambda }_i$ occurs as a root of the characteristic polynomial of $A$

  Let $D=\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]$ and find the geometric and algebraic multiplicity of each eigenvalues of $D$ §

  Computing, $char(D)=(5-\lambda {)}^2$, so 5 is an eigenvalue of $D$ with algebraic multiplicity of 2. The eigenspace of $D$ corresponding to 5 is ${\mathbb{R}}^2$. Thus, the geometric multiplicity of 5 is 2

  Let $J=\left[\begin{array}{cc}5& 1\\ 0& 2\end{array}\right]$ and find the geometric and albegraic multiplicity of each eigenvalues of $J$ §

  Computing $char(J)=(5-\lambda )(2-\lambda )$, so 5 and 2 are eigenvalues of $j$, both with algebraic multiplicity of 1. The eigenspace of $j$ corresponding to 5 is $span\left\{\left[\begin{array}{c}1\\ 0\end{array}\right]\right\}$ and the eigenspace corresponding to 2 is $span\left\{\left[\begin{array}{c}-1\\ 3\end{array}\right]\right\}$. Thus both 5 and 2 has a geometric multiplicity of 1.

• Fundamental Theorem of Allegra - Let $p$ be a polynomial of degree $n$. Then if complex roots are allowed, the sum of multiplicities of the roots of $p$ is $n$

• Let $\lambda$ be an eigenvalues of the matrix $A$. Then

  $$\text{geometric mult}(\lambda )\le \text{algebraic multi }(\lambda )$$
  • An $n\times n$ matrix $A$ is diagonalizable if and only if the sum of its geometric multiplicities is equal to $n$. Further, provided complex eigenvalues are permitted, $A$ is diagonalizable if and only if all its geometric multiplicities equal its algebraic multiplicities.