Transformation of matrices

Unit 7 Transformation of matrices

Key unit competence: Apply matrix transformations into echelon or diagonal form in solving simultaneous linear equations and computing powers of matrices.

Key ideas

• Elementary row operations: swap two rows; multiply a row by a non-zero number; add a multiple of one row to another. They do not change the solutions of a system or the rank.
• Echelon form: each leading (first non-zero) entry is to the right of the one above, and zero rows are at the bottom. In reduced echelon form each leading entry is 1 with zeros above and below it.
• The rank of a matrix is the number of non-zero rows in an echelon form. A system $AX=B$ is consistent iff $\mathrm{rank}A=\mathrm{rank}[A|B]$; it has a unique solution if this rank equals the number of unknowns, otherwise infinitely many.
• For a linear transformation $f(X)=AX$: kernel $\ker f=\{X:AX=\vec{0}\}$; range $\mathrm{Im}f$ = span of the columns of $A$. Rank–nullity: $\dim \ker f+\dim \mathrm{Im}f=n$ (dimension of the domain).
• Eigenvalues $\lambda$ solve $\det (A-\lambda I)=0$; eigenvectors are non-zero solutions of $(A-\lambda I)v=\vec{0}$. Sum of eigenvalues $=\mathrm{tr}A$, product $=\det A$.
• Diagonalization: if $A$ ($n\times n$) has $n$ independent eigenvectors (for example $n$ distinct eigenvalues), $A=PD{P}^{-1}$ with eigenvectors as columns of $P$ and eigenvalues on the diagonal of $D$. Then $A^k=P{D}^k{P}^{-1}$.

Formulas to know

Common misconceptions

• "Multiplying a row by 0 is an elementary operation." No: the multiplier must be non-zero.
• "An eigenvector can be $\vec{0}$." No: eigenvectors are non-zero by definition.
• Putting the eigenvalues in $D$ in a different order from the eigenvectors in $P$.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Solve by Gaussian elimination: $x+y+z=6$, $2x-y+z=3$, $x+2y-z=2$.

   Show answer

   $R_2-2R_1$: $-3y-z=-9$; $R_3-R_1$: $y-2z=-4$. From these $y=2$, $z=3$, then $x=1$. Solution $(1,2,3)$.

   Common mistake: Arithmetic slips when eliminating; always check in all three equations.

2. Given $A=PD{P}^{-1}$ with $D=\mathrm{diag}(1,-1)$, show that $A^2=I$.

   Show answer

   $A^2=P{D}^2{P}^{-1}=PI{P}^{-1}=I$ because $D^2=\mathrm{diag}(1,1)$.

   Common mistake: Writing $A^2=P^2{D}^2$.

3. Find the kernel and the range of $f(x,y,z)=(x-y,y-z,x-z)$.

   Show answer

   $\ker f$: $x=y=z$, so $\ker f=\mathrm{span}\{(1,1,1)\}$ (dim 1). By rank–nullity $\dim \mathrm{Im}f=2$; since $x-z=(x-y)+(y-z)$, $\mathrm{Im}f=\{(a,b,c):c=a+b\}$.

   Common mistake: Claiming the range is ${\mathbb{R}}^3$.

4. Reduce $\left(\begin{array}{ccc}1& 2& 3\\ 2& 4& 7\\ 3& 6& 10\end{array}\right)$ to echelon form and give its rank.

   Show answer

   $R_2-2R_1=(0,0,1)$, $R_3-3R_1=(0,0,1)$, then $R_3-R_2=(0,0,0)$. Echelon form $\left(\begin{array}{ccc}1& 2& 3\\ 0& 0& 1\\ 0& 0& 0\end{array}\right)$: rank 2.

   Common mistake: Counting the zero row.

5. Find the eigenvalues and eigenvectors of $A=\left(\begin{array}{cc}3& 1\\ 0& 2\end{array}\right)$.

   Show answer

   $\lambda =3$: $(A-3I)v=0$ gives $v=(1,0)$. $\lambda =2$: $\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)v=0$ gives $v=(1,-1)$.

   Common mistake: Giving the eigenvector $(0,0)$.

6. Show that $A=\left(\begin{array}{ccc}2& 0& 0\\ 0& 3& 4\\ 0& 4& -3\end{array}\right)$ has eigenvalues 2, 5 and $-5$.

   Show answer

   $\det (A-\lambda I)=(2-\lambda )\left[(3-\lambda )(-3-\lambda )-16\right]=(2-\lambda )({\lambda }^2-25)$. Roots 2, 5, $-5$.

   Common mistake: Expanding along a row with many non-zero entries instead of row 1.

7. For $A=\left(\begin{array}{cc}4& 1\\ 2& 3\end{array}\right)$, find $A^n$.

   Show answer

   $P=\left(\begin{array}{cc}1& 1\\ 1& -2\end{array}\right)$, $D=\mathrm{diag}(5,2)$, $P^{-1}=\frac{1}{3}\left(\begin{array}{cc}2& 1\\ 1& -1\end{array}\right)$. $A^n=\frac{1}{3}\left(\begin{array}{cc}2\cdot 5^n+2^n& 5^n-2^n\\ 2\cdot 5^n-2^{n+1}& 5^n+2^{n+1}\end{array}\right)$ (check $n=1$).

   Common mistake: Computing $P^{-1}DP$ instead of $PD{P}^{-1}$.

8. Discuss, according to $k$, the system $x+y=1$, $x+ky=2$.

   Show answer

   $\det =k-1$. If $k\ne 1$: unique solution $y=\frac{1}{k-1}$, $x=1-\frac{1}{k-1}$. If $k=1$: $x+y=1$ and $x+y=2$, no solution.

   Common mistake: Forgetting the special value $k=1$.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. $A$ has eigenvalues 2 and 5. Find the eigenvalues of $A+3I$.	a. $\left(\begin{array}{cc}7& -2\\ -3& 1\end{array}\right)$
   2. Find the inverse of $\left(\begin{array}{cc}1& 2\\ 3& 7\end{array}\right)$.	b. $32$
   3. $A=PD{P}^{-1}$ with $D=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)$. Find $\det (A^5)$.	c. $\left(\begin{array}{cc}2& -1\\ 1& 3\end{array}\right)$
   4. Find the matrix of $f(x,y)=(2x-y,x+3y)$ in the standard basis.	d. $5$ and $8$

   Show answer

   1 → d ($5$ and $8$) · 2 → a ($\left(\begin{array}{cc}7& -2\\ -3& 1\end{array}\right)$) · 3 → b ($32$) · 4 → c ($\left(\begin{array}{cc}2& -1\\ 1& 3\end{array}\right)$)

   • Step 1. $(A+3I)v=Av+3v=(\lambda +3)v$.
     Step 2. $2+3=5$ and $5+3=8$.
   • Step 1. $\det =7-6=1$.
     Step 2. Swap the diagonal and change the signs of the other two entries; divide by 1.
   • Step 1. $\det A=\det D=1\times 2=2$.
     Step 2. $\det (A^5)=2^5=32$.
   • Step 1. $f(1,0)=(2,1)$ and $f(0,1)=(-1,3)$.
     Step 2. These images are the columns.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $A^2-5A-2I=O$  ·  2  ·  $\mathrm{span}\{(1,-1,1)\}$  ·  $\mathrm{span}\{(1,0,-1)\}$
   1. The rank of $\left(\begin{array}{ccc}1& 2& 3\\ 2& 4& 6\\ 1& 1& 1\end{array}\right)$ is .
   2. The kernel of $f(x,y,z)=(x+y,y+z)$ is .
   3. For $A=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)$, the Cayley–Hamilton theorem gives .
   Show answer

   1: 2 · 2: $\mathrm{span}\{(1,-1,1)\}$ · 3: $A^2-5A-2I=O$

   • Step 1. $R_2=2R_1$, so $R_2-2R_1=(0,0,0)$.
     Step 2. $R_1$ and $R_3$ are not multiples of each other, so the rank is 2.
   • Step 1. $x+y=0$ and $y+z=0$, so $x=-y$ and $z=-y$.
     Step 2. With $y=-1$: $(1,-1,1)$. $\ker f=\mathrm{span}\{(1,-1,1)\}$.
   • Step 1. Characteristic polynomial: ${\lambda }^2-(\text{trace})\lambda +\det ={\lambda }^2-5\lambda -2$.
     Step 2. Every square matrix satisfies its own characteristic equation.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Find an eigenvector of $\left(\begin{array}{cc}4& 1\\ 2& 3\end{array}\right)$ for $\lambda =2$.	a. $\mathrm{span}\{(1,1,1)\}$ and $2$
   2. $A$ has eigenvalues 2 and 5. Find the eigenvalues of $A^{-1}$.	b. $2,3,-1$
   3. Find the eigenvalues of $\left(\begin{array}{ccc}2& 5& 1\\ 0& 3& 4\\ 0& 0& -1\end{array}\right)$.	c. $(1,-2)$
   4. $f:{\mathbb{R}}^3\to {\mathbb{R}}^2$, $f(x,y,z)=(x-y,y-z)$. Find $\ker f$ and $\dim \mathrm{Im}f$.	d. $\frac{1}{2}$ and $\frac{1}{5}$

   Show answer

   1 → c ($(1,-2)$) · 2 → d ($\frac{1}{2}$ and $\frac{1}{5}$) · 3 → b ($2,3,-1$) · 4 → a ($\mathrm{span}\{(1,1,1)\}$ and $2$)

   • Step 1. $A-2I=\left(\begin{array}{cc}2& 1\\ 2& 1\end{array}\right)$, so $2x+y=0$.
     Step 2. $y=-2x$: $(1,-2)$. Check: $A(1,-2)=(2,-4)$.
   • Step 1. $Av=\lambda v\Rightarrow v=\lambda A^{-1}v\Rightarrow A^{-1}v=\frac{1}{\lambda }v$.
     Step 2. So the eigenvalues are $\frac{1}{2}$ and $\frac{1}{5}$.
   • Step 1. The matrix is upper triangular.
     Step 2. The eigenvalues of a triangular matrix are its diagonal entries.
   • Step 1. $x-y=0$ and $y-z=0$ give $x=y=z$: $\ker f=\mathrm{span}\{(1,1,1)\}$.
     Step 2. Rank–nullity: $\dim \mathrm{Im}f=3-1=2$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   1 and 3  ·  infinitely many solutions with one parameter  ·  $\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)$, the reflection in $y=-x$  ·  $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, the reflection in $y=x$
   1. A rotation through $90^{\circ }$ is followed by the reflection in the $x$-axis. The combined matrix is .
   2. Row reduction gives $\left(\begin{array}{cccc}1& 2& 1& 3\\ 0& 1& -1& 1\\ 0& 0& 0& 0\end{array}\right)$. The system has .
   3. The eigenvalues of $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ are .
   Show answer

   1: $\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)$, the reflection in $y=-x$ · 2: infinitely many solutions with one parameter · 3: 1 and 3

   • Step 1. The second map goes on the left: $\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)$.
     Step 2. It sends $(x,y)$ to $(-y,-x)$: the reflection in $y=-x$.
   • Step 1. $\mathrm{rank}A=\mathrm{rank}[A|B]=2$, so it is consistent.
     Step 2. 3 unknowns minus rank 2 gives 1 free parameter.
   • Step 1. $\det (A-\lambda I)=(2-\lambda {)}^2-1=0$.
     Step 2. $2-\lambda =\pm 1$, so $\lambda =1$ or $\lambda =3$.

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