Linear Algebra Checkpoint

Check connections, not just arithmetic

Work these problems without immediately opening the solutions. For each answer, name the space involved and check matrix shapes. A good solution explains existence, uniqueness and sensitivity where relevant, rather than only presenting a number.

The course has two recurring questions: which outputs are reachable, and how can an input be recovered? Elimination answers them algebraically, subspaces organize them geometrically, and QR or SVD supplies useful numerical tools.

Visual guide

VISUAL GUIDESmall singular values amplify inverse sensitivity

Unit circle
Flattened image

The map diag(1, 0.2) flattens the unit circle. Reconstructing an input from its output must multiply the vertical component by 5, including any vertical measurement error. The condition number here is 5.

Problems

Describe every solution of $x+2y+z=3$ , $2x+4y+2z=6$ .
For its $2\times3$ coefficient matrix, give rank, nullity and left nullity.
Project $b=(4,0)^T$ onto the line spanned by $(1,1)^T$ , and verify the residual is orthogonal.
Fit a constant to measurements $1,2,6$ using a normal equation.
Diagonalize $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ and describe $A^k(1,-1)^T$ .
Does $B=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ admit a diagonalization? Explain using its eigenspace.
For $C=\operatorname{diag}(3,0)$ and $b=(6,5)^T$ , find the minimum-norm least-squares solution and residual.
A matrix has singular values $8,2,0.01$ . State its condition number and the spectral error of its best rank-two approximation.

Worked solutions

Show worked solutions

The second equation is redundant. With $y=s,z=t$ , all solutions are $(3-2s-t,s,t)$ . The two free directions form the null space; $(3,0,0)$ is one particular solution.
Rank is one, input nullity is $3-1=2$ , and left nullity is $2-1=1$ . The left-null vector $(-2,1)^T$ gives a consistency test on the right-hand side.
Projection coefficient is $4/2=2$ , so $p=(2,2)^T$ and $r=(2,-2)^T$ . The dot product with $(1,1)^T$ is zero.
The design matrix is a column of three ones. The normal equation is $3c=1+2+6=9$ , giving $c=3$ . Residuals $-2,-1,3$ sum to zero.
Take columns of $V$ as $(1,1)^T,(1,-1)^T$ and $D=\operatorname{diag}(3,1)$ . Then $A=VDV^{-1}$ . The given vector belongs to eigenvalue one, so every power leaves it unchanged.
No. Its only eigenvalue is one and $N(B-I)$ is one-dimensional, so it lacks two independent eigenvectors.
$C^+=\operatorname{diag}(1/3,0)$ , so $x_*=(2,0)^T$ . The fitted vector is $(6,0)^T$ and residual is $(0,5)^T$ . Other minimizing coefficients differ by a vertical null-space vector and have larger norm.
The spectral condition number is $8/0.01=800$ . Discarding the smallest singular contribution gives spectral approximation error $0.01$ .

Where to use these tools next

Differential-equation systems use eigenvectors and matrix dynamics. Multivariable calculus uses gradients, Hessians and quadratic forms. Engineering data fitting uses least squares and conditioning. Return to the relevant unit when a calculation works mechanically but its geometric meaning is still unclear.