MAT223 Lecture 10

• 1

  • We have $A=\left[\begin{array}{ccc}2& 1& 0\\ 0& 0& 0\\ 0& 1& 2\end{array}\right]$
  • $c_A(x)=x(x-2{)}^2$
  • $B=\left[\begin{array}{ccc}2& 0& 0\\ -2& 0& -4\\ 1& 0& 2\end{array}\right]$
  • $c_B(x)=x(x-2{)}^2$
  • We have the same Eigenvalues for both.
  • So we have at most $3$ Eigenvectors but minimum $2$ because we have just $2$ eigenvalues.
  • $A=PD{P}^{-1}$
  • $\left[\begin{array}{ccc}2& 1& 0\\ 0& 0& 0\\ 0& 1& 2\end{array}\right]=\left[\begin{array}{ccc}1& 1& 0\\ -2& 0& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}0& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]P^{-1}$
  • What's $P$ what's $D$?
    • For $D$ and $P$, all you need is the order to be the same between them. Otherwise, you can just do ${\lambda }_0,{\lambda }_2,{\lambda }_1$ or ${\lambda }_2,{\lambda }_1,{\lambda }_0$
    • As long as the order is the same between $D$ and $P$
  • To find Eigenvectors we need to solve Homogeneous Systems.
  • $(A-\lambda I_n)\overrightarrow{x}=\overrightarrow{0}$
  • So $(A-\lambda I_n|0)$ for each value of $\lambda$.

• 4: polls

  • $A=\left[\begin{array}{ccc}0& 0& 1\\ 0& 3& 4\\ 0& 0& -7\end{array}\right]$
  • $c_A(x)=det(\left[\begin{array}{ccc}x& 0& 0\\ 0& x& 0\\ 0& 0& x\end{array}\right]-\left[\begin{array}{ccc}0& 0& 1\\ 0& 3& 4\\ 0& 0& -7\end{array}\right])$
  • $c_A(x)=det\left[\begin{array}{ccc}x& 0& -1\\ 0& x-3& -4\\ 0& 0& x+7\end{array}\right]$
  • $c_A(x)=(x)(x-3)(x+7)=x^3+4x^2-21x$
  • $A-\lambda I_n$ is also upper-triangular.

• 5

  • $\left[\begin{array}{ccc}1& 2& 3\\ 0& -1& 5\\ 0& 0& 1\end{array}\right]$
  • has $x$ distinct eigenvalues
  • $c_A(x)=det\left[\begin{array}{ccc}x-1& -2& -3\\ 0& x+1& \end{array}\right]$
  • …
  • Since we know empirically, we can just say that eventually once you do the above you get $3$ values but you get $\lambda =1,-1$

• 6

  • Suppose that $A$ is $3\times 3$
  • We have $\lambda =1,-2$
  • What could be the $c_A(x)$?
  • $c_A(x)=(x-1)(x+2{)}^2$
  • $c_A(x)=(x-1{)}^2(x+2)$

• 7

  • Is it true or false
  • $3\times 3$ matrix $\lambda =1,3,-4$
  • Then it must be diagonalizable
  • Yes because $3$ distinct $\lambda$ values results in the minimum of $3$ basic eigenvectors which matches the dimensions of the original.

• 2

  • We have $\left[\begin{array}{ccc}2& 1& 0\\ 0& 0& 0\\ 0& 1& 2\end{array}\right]$ and $\left[\begin{array}{ccc}2& 0& 0\\ -2& 0& -4\\ 1& 0& 2\end{array}\right]$
  • First has a row of $0$s so it's not invertible.
  • Second has a column of $0$s so it's not invertible.
  • You're invertible exactly when $\mathrm{rank}\left(A\right)=n$ and it's $n\times n$
  • They both have rank of $2$
  • If $C$ is square
    • If it's not diagonalizable then $C$ is not invertible
      • $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$
      • $detC=1$
      • $\left[\begin{array}{cc}x-1& 1\\ 0& x-1\end{array}\right]$
      • $(x-1{)}^2$
      • $\lambda =1$
      • $\left[\begin{array}{ccc}x-1& 1& 0\\ 0& x-1& 0\end{array}\right]$
      • $\left[\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\end{array}\right]$
      • Inconsistent.
      • Something invertible but not diagonalizable
        • $I_n$
        • $detC=1$ and $\mathrm{rank}\left(C\right)=2$
        • To show it's not diagonalizable
        • $C_C(x)=(x-1{)}^2$
        • Solve the Homogeneous System and find one Basic Eigenvector
      • $C=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$ has no real eigenvalues.
        • $C_C(x)=x^2+1$
        • Invertible not diagonalizable
    • Vice versa
      • If it's not invertible, that means $detC=0$
      • So there wouldn't be enough eigenvalues.
      • $det(C-\lambda I_n)$
      • Non-invertible, is diagonalizable
      • $A=\left[\begin{array}{ccc}2& 1& 0\\ 0& 0& 0\\ 0& 1& 2\end{array}\right]$
      • $detA=0$ not invertible
      • But it is diagonalizable
    • There is no connection between diag and invertibility.

• 3

  • $C$ is square
  • Show these are true
  • 1: if $C$ is non-invertible. Then $0$ is an eigenvalue of $C$
    • We have a square $C$. It's not invertible meaning that $detC=0$
    • Because $detC=0$. This means that $\lambda I_n-A$ will have a $0$ eigenvalue due to a $0$ determinant.
    • If it's invertible it has only the trivial solution. So there are no Eigenvectors for $\lambda =0$
    • Otherwise, they'd be non-trivial solutions to $C\overrightarrow{x}=\overrightarrow{0}$.
  • 2: if $C$ has $0$ as an eigenvalue, then $C$ is non-invertible.
    • We have $c_C(x)=x(x+{\lambda }_1)(x+{\lambda }_2)\dots$
    • The only way to get $\lambda =0$ is by having $detC=0$
    • The reverse direction holds.
    • #tk idk the above, doesn't make sense.
  • 3:

• Polls:

  • If we have $A=3\times 4$
  • $B=4\times 5$
  • $C=2\times 5$
  • $\overrightarrow{x}=5\times 1$
  • $BA=4\times 5\cdot 3\times 4$
    • Not defined
  • $AB=3\times 4\cdot 4\times 5$
    • This is defined
  • $(B{C}^T{)}^T$
    • $(4\times 5\cdot 5\times 2)$
    • Defined
  • $A+B$
    • $3\times 4+4\times 5$
    • not
  • $A\overrightarrow{x}$
    • $3\times 4\cdot 5\times 1$
    • Not
  • $C\overrightarrow{x}$
    • $2\times 5\cdot 5\times 1$
  • 2
    • $A$ is $n\times n$
    • Which of the following are equivalent to $A$ is invertible
    • $detA=0$
      • no
    • $A\overrightarrow{x}=\overrightarrow{0}$ has only trivial solution
      • Meaning rank is $<n$
      • Row of $0$s
      • Not intervible
    • the RREF of $A$ is $I_n$
    • If we have $A$ and $\mathrm{rank}\left(A\right)<n$ then $detA=0$

• $A\overrightarrow{x}=\overrightarrow{0}$

  • If it's invertible, it's only trivial solutions.
  • If it's non-invertible, then it has infinite solutions.

• Three possibilities for any solutions.

  • If invertible, then only 1 sol
  • If it's invertible, rank $A=n$
    • So our remaining parameters are $n-r=0$
    • So that means only $1$ solution.