Try each problem before expanding its solution. Check answers by substitution or multiplication.
1. An affine solution plane
Solve $x+2y-z=3$ and $2x+4y-2z=6$.
Worked solutionThe second equation is redundant. Set $y=s,z=t$, giving
$$(x,y,z)=(3-2s+t,s,t)=(3,0,0)+s(-2,1,0)+t(1,0,1).$$
The coefficient matrix has rank one and nullity two. The nonhomogeneous solution set is an affine plane, not a subspace.
2. Test a basis
Do $(1,0,1)$, $(0,1,1)$, $(1,1,2)$ form a basis of $\mathbb R^3$?
Worked solutionNo: the third vector is the sum of the first two. The first two are independent and form a basis of the plane $z=x+y$. The three-column matrix has rank two and a null vector $(-1,-1,1)^T$.
3. Change coordinates
For $B=((1,0),(1,1))$, find $[(5,3)]_B$.
Worked solutionWrite $(5,3)=a(1,0)+b(1,1)=(a+b,b)$. Hence $b=3,a=2$ and the coordinate vector is $(2,3)^T$. The basis matrix $\begin{bmatrix}1&1\\0&1\end{bmatrix}$ converts these coordinates back to $(5,3)^T$.
4. Project onto a line
Project $v=(2,0)^T$ onto the span of $u=(1,1)^T$.
Worked solution$$p=u\frac{u^Tv}{u^Tu}=(1,1)^T.$$
The residual is $(1,-1)^T$ and its dot product with $u$ is zero. The projection matrix is $\frac12\begin{bmatrix}1&1\\1&1\end{bmatrix}$.
5. Least-squares constant
Fit one constant to observations $1,2,4$.
Worked solutionSet $A=(1,1,1)^T$, $b=(1,2,4)^T$. The normal equation is $3c=7$, so $\hat c=7/3$. The residual is $(-4/3,-1/3,5/3)^T$, sums to zero, and has squared norm $14/3$.
6. Diagonalize or explain failure
Compare $A=\begin{bmatrix}2&1\\1&2\end{bmatrix}$ and $J=\begin{bmatrix}1&1\\0&1\end{bmatrix}$.
Worked solutionFor $A$, eigenvalues are $3,1$, with eigenvectors $(1,1)^T,(1,-1)^T$. Normalize them as columns of $Q$ to obtain $A=Q\operatorname{diag}(3,1)Q^T$.
For $J$, the sole eigenvalue $1$ has algebraic multiplicity two but eigenspace $\operatorname{span}\{(1,0)^T\}$ of dimension one. It is not diagonalizable despite its characteristic polynomial splitting.
7. A small determinant
Does $\det A$ being tiny imply ill-conditioning when $A=10^{-8}I_3$?
Worked solutionNo. The determinant is $10^{-24}$, but all singular values equal $10^{-8}$, so $\kappa_2(A)=1$. Compare $\operatorname{diag}(1,1,10^{-8})$, which has condition number $10^8$.
8. Complex inner product
For $x=(1,i)^T$ and $y=(i,1)^T$, compute $\langle x,y\rangle$ and both norms.
Worked solutionUsing $\langle x,y\rangle=x^*y$, the value is $i-i=0$. Both norms are $\sqrt2$, so the vectors are orthogonal. Using $x^Ty$ would incorrectly give $2i$.
9. Markov orientation
For row-stochastic $P=\begin{bmatrix}0.9&0.1\\0.5&0.5\end{bmatrix}$ and initial column state $q_0=(1,0)^T$, what is $q_1$?
Worked solutionUse $q_1=P^Tq_0=(0.9,0.1)^T$. The wrong formula $Pq_0$ gives $(0.9,0.5)^T$, whose entries sum to $1.4$ and thus do not form a probability distribution. The stationary column state is $(5/6,1/6)^T$.
10. Generalized kernel and range
Let $T=\operatorname{diag}(J_2(0),2)$ on $\mathbb R^3$, where $J_2(0)=\begin{bmatrix}0&1\\0&0\end{bmatrix}$. Find the stabilized kernel and image.
Worked solution$T^2=\operatorname{diag}(0,0,4)$. The stabilized kernel is $\operatorname{span}\{e_1,e_2\}$, and the stabilized image is $\operatorname{span}\{e_3\}$. They form a direct sum of the whole space. $T$ is nilpotent on the first part and invertible on the second.
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