[MEEG690]

Exam 1 (Solution)

1.

a) (3pts) The linear system \(A \boldsymbol{x}= \boldsymbol{b}\) has at least one solution if and only if it is consistent (compatible), i.e. vector \(\boldsymbol{b}\) lies in the image space of \(A\) (or if and only if the reduced row-echelon form of \((A|\boldsymbol{b})\) does not contain an inconsistent row).

b) (3pts) The compatible linear system \(A \boldsymbol{x}= \boldsymbol{b}\) admits a unique solution if and only if the sole element of the kernel of \(A\) is \(\boldsymbol{z}=\boldsymbol{0}\).

c) (3pts) The kernel of \(A\) is the orthogonal complement of the image of \(A^T\) (coimage) in \(\mathbb{R}^n\). The image of \(A\) is the orthogonal complement of the kernel of \(A^T\) (cokernel) in \(\mathbb{R}^m\).

2.

a) (20 pts) First we find matrix \(A\): \[ A= \boldsymbol{u}\boldsymbol{v}^T= \begin{pmatrix} 1 \\ 2 \end{pmatrix} \begin{pmatrix} -1 & 3 \end{pmatrix} = \begin{pmatrix} -1 & 3 \\ -2 & 6 \end{pmatrix}= (-\boldsymbol{u}|3\boldsymbol{u}) \] Also note that \(A^T= (\boldsymbol{v}| 2\boldsymbol{v})\).

  1. Find \(\ker A\): We solve \(A\boldsymbol{z}=\boldsymbol{0}\) that is \((-z_1+3z_2 )\boldsymbol{u}= \boldsymbol{0}\) giving \(-z_1+3z_2=0\). We conclude that \[ \boxed{ \ker A= \text{span}( \boldsymbol{v}^\perp )}, \qquad \boldsymbol{v}^\perp = \begin{pmatrix} 3 \\ 1 \end{pmatrix} \] Note that \(\boldsymbol{v}^\perp \cdot \boldsymbol{v}=0\).

  2. Find \(\text{img}A\), that is, the set of \(\boldsymbol{y}\in\mathbb{R}^2\) such that \(\boldsymbol{y}= A\boldsymbol{x}= (-x_1+3x_2)\boldsymbol{u}\). Hence \[ \boxed{ \text{img}A=\text{span}(\boldsymbol{u})} \]

  3. Find \(\ker A^T\): We solve \(A^T\boldsymbol{z}'=\boldsymbol{0}\) that is \((z_1'+2z'_2 )\boldsymbol{v}= \boldsymbol{0}\) giving \(z'_1+2z'_2=0\). We conclude that \[ \boxed{ \ker A^T = \text{span}(\boldsymbol{u}^\perp)}, \qquad \boldsymbol{u}^\perp = \begin{pmatrix} -2 \\ 1 \end{pmatrix} \] Note that \(\boldsymbol{u}^\perp \cdot \boldsymbol{u}=0\).

  4. Find \(\text{img}A^T\), that is, the set of \(\boldsymbol{y}'\in\mathbb{R}^2\) such that \(\boldsymbol{y}' = A^T\boldsymbol{x}'= (x'_1+2x'_2)\boldsymbol{v}\). Hence \[ \boxed{ \text{img}A^T =\text{span}(\boldsymbol{v})} \]

We have already noted the orthogonality of the spaces \(\ker A = \text{span}(\boldsymbol{v}^\perp)\) and \(\text{img}A^T= \text{span}(\boldsymbol{v})\) since \(\boldsymbol{v}\cdot\boldsymbol{v}^\perp=0\): \(\ker A = (\text{img}A^T)^\perp\). Likewise, the orthogonality of the spaces \(\ker A^T = \text{span}(\boldsymbol{u}^\perp)\) and \(\text{img}A= \text{span}(\boldsymbol{u})\) since \(\boldsymbol{u}\cdot\boldsymbol{u}^\perp=0\): \(\ker A^T = (\text{img}A)^\perp\). We can sketch \(\ker A\) and \(\text{img}A^T\) in the domain \(\mathbb{R}^2\) of \(A\) as two orthogonal lines of the plane.

Likewise, we can sketch \(\ker A^T\) and \(\text{img}A\) in the codomain \(\mathbb{R}^2\) of \(A\) as two orthogonal lines of the plane.

b) (5 pts) The compatibility condition satisfied by vector \(\boldsymbol{b}\) to guarantee that the linear system \(A\boldsymbol{x}=\boldsymbol{b}\) has at least one solution is that \(\boldsymbol{b}\in \text{img}(A)= \text{span}(\boldsymbol{u})\) or equivalently \(\boldsymbol{b}\cdot \boldsymbol{u}^\perp =0\): \[ \boxed{-2b_1 + b_2 = 0} \]

c) (20 pts) The general solution of \(A\boldsymbol{x}=\boldsymbol{b}\) is of the form \(\boldsymbol{x}=\boldsymbol{w}+\boldsymbol{z}\) where \(\boldsymbol{z}\in \ker A= \lambda\boldsymbol{v}^\perp\), and where \(\boldsymbol{w}\) is the particular solution of minimum Euclidean norm. We know that \(A\boldsymbol{w}=\boldsymbol{b}\) and \(\boldsymbol{w}\in \text{img}A^T\). Since \(\text{img}A^T = (\ker A)^\perp\) we require \(\boldsymbol{w}\cdot \boldsymbol{v}^\perp=0\): \[ A\boldsymbol{w}= \boldsymbol{b}, \quad \text{ and } 3w_1 + w_2 =0 \] This implies that we must examine the solution \(\boldsymbol{w}\) of the augmented system \[ \begin{pmatrix} -1 & 3 & b_1 \\ -2 & 6 & b_2 \\ 3 & 1 & 0 \end{pmatrix} \Longrightarrow \begin{pmatrix} -1 & 3 & b_1 \\ 0 & 0 & -2b_1+ b_2 \\ 0 & 10 & 3b_1 \end{pmatrix} \] So if \(b_2 -2 b_1 =0\) (the compatibility condition satisfied by \(\boldsymbol{b}\)), then vector \(\boldsymbol{w}\) is given by \[ \boxed{w_2 = \frac{3}{10} b_1 \Longrightarrow w_1 = -\frac{1}{10} b_1} \] We conclude that the general solution of \(A\boldsymbol{x}=\boldsymbol{b}\) is of the form \[ \boxed{ \boldsymbol{x}= \frac{b_1}{10} \begin{pmatrix} -1 \\ 3 \end{pmatrix} +\lambda \begin{pmatrix} 3 \\ 1 \end{pmatrix} } \]

3.

a) (15 pts) To show that \(\langle p, q\rangle\) defines an inner product on \({\cal P}_2\), we need to show:

  1. symmetry: given \(p, q\in {\cal P}_2\) \[ \langle p, q\rangle= p(-1) q(-1) + p(0)q(0) + p(1) q(1)= q(-1) p(-1) + q(0)p(0) + q(1) p(1)= \langle q, p\rangle \]
  2. linearity (w.r.t first argument): given \(p_1, p_2 , q\in {\cal P}_2\) and \(a,b\in\mathbb{R}\) \[ \langle (ap_1+ bp_2), q\rangle= (a p_1 (-1) +b p_2 (1)) q(-1) + (ap_1 (0)+ bp_2 (0)) q(0) + (a p_1(1)+b p_2(1)) q(1) \] \[ \langle (ap_1+ bp_2), q\rangle= = a (p_1(-1) q(-1) + p_1(0)q(0) + p_1(1) q(1)) +b (p_2(-1) q(-1) + p_2(0)q(0) + p_2(1) q(1)) = a\langle p_1, q\rangle+ b\langle p_2, q\rangle \]
  3. positive definite: given \(p\in {\cal P}_2\), \[ \langle p, p \rangle= p(-1)^2 + p(0)^2 + p(1)^2 \geq 0 \] Also \[ \langle p, p \rangle=0 \Longrightarrow p(-1)^2 + p(0)^2 + p(1)^2 =0 \Longrightarrow p(-1)= p(0)= p(1) =0 \] This would imply that quadratic polynomial \(p\) has 3 distinct roots: this is only possible if \(p=0\).

b) (15 pts) To find an orthonormal basis of \({\cal P}_2\) for this inner product, we find 3 distinct quadratic polynomials which have 2 roots among \(x= -1, 0, 1\): this gives (within a multiplicative constant) \[ p_0 (x) = (1-x)(1+x) = 1-x^2 \] \[ p_1(x)= x(x+1) \] \[ p_2 (x) = x(x-1) \] Clearly \(\langle p_0, p_1 \rangle = p_0(-1) p_1(-1) + p_0(0)p_1(0) + p_0(1) p_1(1)=0\) since \(p_0 (\pm 1)=0\) and \(p_1 (-1)= p_1 (0)=0\). For the same reason, \(\langle p_0, p_2 \rangle = 0\) and \(\langle p_1, p_2 \rangle = 0\) So \((p_0,p_1,p_2)\) is an orthogonal set, and thus necessarily linear independent. The set \((p_0,p_1,p_2)\) defines an orthogonal basis of \({\cal P}_2\).

We need to normalize \((p_0,p_1,p_2)\): we define \(q_k = p_k / \| p_k \|\): with \(\|p_0\|=1\) and \(\|p_1\|=\|p_2\|=2\) \[ \boxed{ q_0 (x) = p_0 (x) = 1-x^2 } \] \[ \boxed{ q_1 (x) = \frac{1}{2} p_1 (x) =\frac{1}{2} x(x+1) } \] \[ \boxed{ q_2 (x) = \frac{1}{2} p_2 (x) =\frac{1}{2} x(x-1) } \] Now \((q_0,q_1,q_2)\) defines an orthonormal basis of \({\cal P}_2\).

c) (15 pts) Let \(U= \{p\in{\cal P}_2 \text{ such that } p(-1)+p(0)+p(1)=0\}\). We need find a basis of \(U\): we seek \(p = c_0 q_0 +c_1 q_1 + c_2 q_2\) (a linear combination of \(q_0\), \(q_1\) and \(q_2\)) and impose \(p(-1)+p(0)+p(1)=0\). This gives \[ (0 c_0 + 0 c_1 + c_2) + (c_0 + 0 c_1 + 0c_2 ) + (0 c_0 + c_1 + 0 c_2 ) =0 \] leading to \[ c_0 + c_1 + c_2 =0 \Longrightarrow c_2 = -c_0 -c_1 \] This shows that \(p\in U\) takes the form \[ p= c_0 (q_0 -q_2) + c_1 (q_1 -q_2) \] for any \(c_0, c_1 \in\mathbb{R}\) and conclude that \(\boxed{U= \text{span}(q_0-q_2, q_1 -q_2)}\).

We find \(U^\perp\) by following the same steps: \(p \in U^\perp\) if and only if \[ \langle p, q_0-q_2 \rangle = \langle p, q_1-q_2 \rangle =0 \] Let \(p= c_0 q_0 +c_1 q_1 + c_2 q_2 \in U^\perp\): we find \[ \langle p, q_0-q_2 \rangle = \langle c_0 q_0 +c_1 q_1 + c_2 q_2, q_0-q_2 \rangle= c_0 \langle q_0, q_0 \rangle - c_2 \langle q_2, q_2 \rangle= c_0 -c_2 \] \[ \langle p, q_1-q_2 \rangle = \langle c_0 q_0 +c_1 q_1 + c_2 q_2, q_1-q_2 \rangle= c_1 \langle q_1, q_1 \rangle - c_2 \langle q_2, q_2 \rangle= c_1 -c_2 \] This shows that \(c_0 = c_1 = c_2\): \[ p= c_0 (q_0+q_1 + q_2) \Longrightarrow \boxed{U^\perp=\text{span}(q_0+q_1+q_2)= \text{span}(1)} \] since \(q_0+q_1+q_2 =1\).