Phase 5 Exam
Numerical Linear Algebra & Computing - Phase Exam
15 interleaved questions drawn from every lesson in this phase. Interleaving mixes topics on purpose - that difficulty is what builds durable, transferable understanding. Aim for 70%+ before advancing; below that, revisit the flagged lessons.
How this exam teachesQuestions are shuffled across lessons (not blocked by topic) so you practice choosing the right idea. Missed questions are added to your review queue automatically.
Q1 MediumA matrix has condition number \(\kappa\approx10^{8}\). Using a backward-stable double-precision solver, roughly how many correct digits can you expect?
About 16
About 8
About 0
It depends only on the algorithm, not \(\kappa\)
Q2 MediumWhy does partial pivoting improve numerical stability?
It reduces the flop count
It keeps every multiplier \(|\ell_{ik}|\le1\), bounding error growth
It makes \(A\) symmetric
It removes the need for back substitution
Q3 MediumLeast squares via \(Rx=Q^\top b\) is preferred over the normal equations mainly because:
It uses fewer flops
It keeps the conditioning at \(\kappa(A)\) instead of \(\kappa(A)^2\)
It works for singular \(A\)
It avoids computing \(Q\)
Q4 MediumCholesky needs no pivoting because:
It only works on triangular matrices
Positive-definiteness keeps every pivot (diagonal radicand) strictly positive
It ignores rounding error
It uses complex arithmetic
Q5 MediumWhy must eigenvalue algorithms be iterative rather than closed-form?
Because matrices are too big
Because eigenvalues are roots of a polynomial, and there is no general formula past degree 4 (Abel–Ruffini)
Because of floating-point error only
Because eigenvalues are always complex
Q6 EasyWhich statement best separates conditioning from stability?
Both are properties of the algorithm
Conditioning is a property of the problem; stability is a property of the algorithm
Conditioning is a property of the algorithm; stability of the problem
They are two names for the same thing
Q7 MediumYou must solve \(Ax=b_i\) for 100 different \(b_i\) with the same \(A\). The efficient approach is:
Compute \(A^{-1}\) once, multiply 100 times
Factor \(PA=LU\) once, then 100 pairs of triangular solves
Run full Gaussian elimination 100 times
Use Cramer’s rule per system
Q8 MediumThe single algorithmic difference between classical and modified Gram-Schmidt is:
MGS normalizes twice
MGS subtracts projections against the running, partially-orthogonalized vector rather than the original column
MGS uses a different \(R\)
MGS requires pivoting
Q9 MediumTo draw \(X\sim\Normal(0,\Sigma)\) from standard normals \(Z\) with \(\Sigma=LL^\top\), set:
\(X=\Sigma Z\)
\(X=LZ\)
\(X=L^{-1}Z\)
\(X=Z/L\)
Q10 MediumPower iteration converges to the dominant eigenvector at a linear rate equal to:
\(|\lambda_1|\)
\(|\lambda_2/\lambda_1|\)
\(|\lambda_1/\lambda_2|\)
\(\kappa(A)\)
Q11 MediumCatastrophic cancellation is most severe when you:
Add two large numbers of the same sign
Multiply two nearly equal numbers
Subtract two nearly equal, already-rounded numbers
Divide by a number near 1
Q12 EasyIn \(PA=LU\) with partial pivoting, the matrix \(L\) is:
Upper-triangular with unit diagonal
Unit lower-triangular with entries \(|\ell_{ik}|\le1\)
Orthogonal
Diagonal
Q13 EasyFor a full-rank \(A=QR\) (reduced), the least-squares residual \(b-Ax^\star\) is:
Zero
Orthogonal to the columns of \(A\)
Parallel to \(b\)
Equal to \(Q^\top b\)
Q14 EasyA failed Cholesky (a non-positive radicand) on an estimated covariance most likely means:
The code has a bug for sure
The matrix is not positive definite - often too few samples or a bad estimate
The matrix is too large
You must switch to LU
Q15 EasyFor a very large sparse symmetric positive-definite system, the method of choice is:
Dense LU
Explicit matrix inverse
Conjugate gradients (a Krylov method)
Cramer’s rule