Homework 3

1. Find the matrix representation for the following linear transformations.

(a) T(x1,x2,x3)=(3x1−x2,x2+x3,x1−x2−x3)T(x_1, x_2, x_3) = (3x_1 - x_2, x_2 + x_3, x_1 - x_2 - x_3)

T:R3→R3T(x⃗)=(3−100111−1−1)(x1x2x3)T: \R^3 \to \R^3 \\ T(\vec{x}) = \begin{pmatrix} 3 & -1 & 0 \\ 0 & 1 & 1 \\ 1 & -1 & -1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}

(b) TT maps [100]\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix}, and[010]\begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix} and[001]\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix} respectively to[01]\begin{bmatrix} 0 \\ 1\end{bmatrix} ,[11]\begin{bmatrix} 1 \\ 1\end{bmatrix} ,[1−1]\begin{bmatrix} 1 \\ -1\end{bmatrix}

T:R3→R2T(x⃗)=(01111−1)(x1x2x3)T: \R^3 \to \R^2 \\ T(\vec{x}) = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 1 & -1 \\ \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}

(c) T(x1,x2)=x1[112]+x2[−115]T(x_1, x_2) = x_1 \begin{bmatrix} 1 \\ 1 \\ 2\end{bmatrix} + x_2 \begin{bmatrix} -1 \\ 1 \\ 5\end{bmatrix}

T:R2→R3T(x⃗)=(1−11125)(x1x2)T: \R^2 \to \R^3 \\ T(\vec{x}) = \begin{pmatrix} 1 & -1 \\ 1 & 1 \\ 2 & 5 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}

2. Sketch the image of the square formed by vertices (0,0)(0, 0), (0,1)(0, 1), (1,0)(1, 0) and (1,1)(1, 1) under the linear transformation T(x)=[1−123]xT(\mathbf{x}) = \begin{bmatrix} 1 & -1 \\ 2 & 3\end{bmatrix}\mathbf{x}.

The transformed vertices and its image are in red.

T[00]=[1−123][00]=[1(0)+(−1)(0)2(0)+3(0)]=[00]T[01]=[1−123][01]=[1(0)+(−1)(1)2(0)+3(1)]=[−13]T[10]=[1−123][10]=[1(1)+(−1)(0)2(1)+3(0)]=[12]T[11]=[1−123][11]=[1(1)+(−1)(1)2(1)+3(1)]=[05]\begin{array}{c} T\begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \end{bmatrix} &= &\begin{bmatrix} 1(0) + (-1)(0) \\ 2(0) + 3(0) \end{bmatrix} &= &\begin{bmatrix} 0 \\ 0 \end{bmatrix} \\ T\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} &= &\begin{bmatrix} 1(0) + (-1)(1) \\ 2(0) + 3(1) \end{bmatrix} &= &\begin{bmatrix} -1 \\ 3 \end{bmatrix} \\ T\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} &= &\begin{bmatrix} 1(1) + (-1)(0) \\ 2(1) + 3(0) \end{bmatrix} &= &\begin{bmatrix} 1 \\ 2 \end{bmatrix} \\ T\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} &= &\begin{bmatrix} 1(1) + (-1)(1) \\ 2(1) + 3(1) \end{bmatrix} &= &\begin{bmatrix} 0 \\ 5 \end{bmatrix} \end{array}

3. Sketch the image of the triangle formed by vertices (−1,1)(-1, 1), (1,0)(1, 0) and (1,1)(1, 1) under the linear transformation T(x)=[4−101]xT(\mathbf{x}) = \begin{bmatrix} 4 & -1 \\ 0 & 1\end{bmatrix}\mathbf{x}.

The transformed vertices and its image are in red.

T[−11]=[4−101][−11]=[4(−1)+(−1)(1)0(−1)+1(1)]=[−51]T[10]=[4−101][10]=[4(1)+(−1)(0)0(1)+1(0)]=[40]T[11]=[4−101][11]=[4(1)+(−1)(1)0(1)+1(1)]=[31]\begin{array}{c} T\begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 4 & -1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 \\ 1 \end{bmatrix} &= &\begin{bmatrix} 4(-1) + (-1)(1) \\ 0(-1) + 1(1) \end{bmatrix} &= &\begin{bmatrix} -5 \\ 1 \end{bmatrix} \\ T\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 4 & -1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} &= &\begin{bmatrix} 4(1) + (-1)(0) \\ 0(1) + 1(0) \end{bmatrix} &= &\begin{bmatrix} 4 \\ 0 \end{bmatrix} \\ T\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 4 & -1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} &= &\begin{bmatrix} 4(1) + (-1)(1) \\ 0(1) + 1(1) \end{bmatrix} &= &\begin{bmatrix} 3 \\ 1 \end{bmatrix} \end{array}

4. Let A=[124−3]A=\begin{bmatrix} 1 & 2 \\ 4 & -3\end{bmatrix}.

(a) Find A2A^2 and A3A^3.

Since AA is a 2×22\times 2 square matrix, AkA^k is well defined for all natural kk.

A2=[124−3][124−3]=[1(1)+2(4)1(2)+2(−3)4(1)+(−3)(4)4(2)+(−3)(−3)]=[9−4−817]A3=[9−4−817][124−3]=[9(1)+(−4)(4)9(2)+(−4)(−3)−8(1)+17(4)−8(2)+17(−3)]=[−73060−67]\begin{align*} A^2 &= \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} \\ &= \begin{bmatrix} 1(1) + 2(4) & 1(2) + 2(-3) \\ 4(1) + (-3)(4) & 4(2) + (-3)(-3) \end{bmatrix} \\ &= \begin{bmatrix} 9 & -4 \\ -8 & 17 \end{bmatrix} \end{align*} \\[2em] \begin{align*} A^3 &= \begin{bmatrix} 9 & -4 \\ -8 & 17 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} \\ &= \begin{bmatrix} 9(1) + (-4)(4) & 9(2) + (-4)(-3) \\ -8(1) + 17(4) & -8(2) + 17(-3) \end{bmatrix} \\ &= \begin{bmatrix} -7 & 30 \\ 60 & -67 \end{bmatrix} \end{align*}

(b) Find 2A3−4A+5I22A^3 - 4A + 5I_2 and A2+2A−11I2A^2 + 2A - 11I_2.

Assuming I2=[1001]I_2 = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}.

2A3−4A+5I2=2[124−3]3−4[124−3]+5[1001]=[−1460120−134]−[4816−12]+[5005]=[−1352104−117]A2+2A−11I2=[124−3]2+2[124−3]−11[1001]=[9−4−817]+[248−6]−[110011]=[0000]\begin{align*} 2A^3 - 4A + 5I_2 &= 2 \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix}^3 - 4 \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} + 5 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} -14 & 60 \\ 120 & -134 \end{bmatrix} - \begin{bmatrix} 4 & 8 \\ 16 & -12 \end{bmatrix} + \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \\ &= \begin{bmatrix} -13 & 52 \\ 104 & -117 \end{bmatrix} \end{align*} \\[2em] \begin{align*} A^2 + 2A - 11I_2 &= \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix}^2 + 2 \begin{bmatrix} 1 & 2 \\ 4 & -3 \end{bmatrix} - 11 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 9 & -4 \\ -8 & 17 \end{bmatrix} + \begin{bmatrix} 2 & 4 \\ 8 & -6 \end{bmatrix} - \begin{bmatrix} 11 & 0 \\ 0 & 11 \end{bmatrix} \\ &= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \end{align*}

5. Let v=[v1v2v3]\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \\ v_3\end{bmatrix}, vT=[v1v2v3]\mathbf{v}^T = \begin{bmatrix} v_1 & v_2 & v_3\end{bmatrix}, and v1≠0v_1 \neq 0.

(a) Find vTv\mathbf{v}^T\mathbf{v} and vvT\mathbf{vv}^T.

vTv=[v1v2v3][v1v2v3]=[v12+v22+v32]vvT=[v1v2v3][v1v2v3]=[v12v1v2v1v3v1v2v22v2v3v1v3v2v3v32]\begin{align*} \mathbf{v}^T\mathbf{v} &= \begin{bmatrix} v_1 & v_2 & v_3 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} \\ &= \begin{bmatrix} v_1^2 + v_2^2 + v_3^2 \end{bmatrix} \end{align*} \\[2em] \begin{align*} \mathbf{vv}^T &= \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} \begin{bmatrix} v_1 & v_2 & v_3 \end{bmatrix} \\ &= \begin{bmatrix} v_1^2 & v_1 v_2 & v_1 v_3 \\ v_1v_2 & v_2^2 & v_2v_3 \\ v_1v_3 & v_2v_3 & v_3^2 \end{bmatrix} \end{align*}

(b) If v≠0\mathbf{v}\neq\mathbf{0}, verify that the rank of vTv\mathbf{v}^T\mathbf{v} and vvT\mathbf{vv}^T are 11.

v1≠0  ⟹  v12>0  ⟺  v12+v22+v32>0∴v≠0∧v1≠0  ⟹  vTv≠0  ⟺  rank⁡(vTv)=1vvT=[v12v1v2v1v3v1v2v22v2v3v1v3v2v3v32]→1v3R31v2R2[v12v1v2v1v3v1v2v3v1v2v3]→R3−R2R2−R3[v12v1v2v1v3000000]∴v≠0∧v1≠0  ⟹  vvT≠0  ⟺  rank⁡(vvT)=1\begin{align*} v_1\neq0 &\implies v_1^2 > 0 \\ &\iff v_1^2 + v_2^2 + v_3^2 > 0 \end{align*} \\ \therefore \mathbf{v}\neq\mathbf{0} \land v_1\neq 0 \implies \mathbf{v}^T\mathbf{v} \neq \mathbf{0} \iff \operatorname{rank}(\mathbf{v}^T\mathbf{v}) = 1 \\[2em] \begin{array}{c} \mathbf{vv}^T = \begin{bmatrix} v_1^2 & v_1 v_2 & v_1 v_3 \\ v_1v_2 & v_2^2 & v_2v_3 \\ v_1v_3 & v_2v_3 & v_3^2 \end{bmatrix} &\xrightarrow[\frac{1}{v_3} R_3]{\frac{1}{v_2} R_2} &\begin{bmatrix} v_1^2 & v_1 v_2 & v_1 v_3 \\ v_1 & v_2 & v_3 \\ v_1 & v_2 & v_3 \end{bmatrix} \\ &\xrightarrow[R_3 - R_2]{R_2 - R_3} &\begin{bmatrix} v_1^2 & v_1 v_2 & v_1 v_3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \end{array} \\ \therefore \mathbf{v}\neq\mathbf{0} \land v_1\neq 0 \implies \mathbf{vv}^T \neq \mathbf{0} \iff \operatorname{rank}(\mathbf{vv}^T) = 1

Homework 3
  1. Find the matrix representation for the following linear transformations.
  1. Sketch the image of the square formed by vertices (0,0)(0, 0), (0,1)(0, 1), (1,0)(1, 0) and (1,1)(1, 1) under the linear transformation T(x)=[1−123]xT(\mathbf{x}) = \begin{bmatrix} 1 & -1 \\ 2 & 3\end{bmatrix}\mathbf{x}.
  1. Sketch the image of the triangle formed by vertices (−1,1)(-1, 1), (1,0)(1, 0) and (1,1)(1, 1) under the linear transformation T(x)=[4−101]xT(\mathbf{x}) = \begin{bmatrix} 4 & -1 \\ 0 & 1\end{bmatrix}\mathbf{x}.
  1. Let A=[124−3]A=\begin{bmatrix} 1 & 2 \\ 4 & -3\end{bmatrix}.
  1. Let v=[v1v2v3]\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \\ v_3\end{bmatrix}, vT=[v1v2v3]\mathbf{v}^T = \begin{bmatrix} v_1 & v_2 & v_3\end{bmatrix}, and v1≠0v_1 \neq 0.
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