University Linear Algebra — Rank and the Fundamental Subspaces Worksheet

All 11 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1Solution Sets in Parametric Vector Form

   The augmented matrix of a linear system in the unknowns $x_1,x_2,x_3,x_4,x_5$ has been row reduced to $$\left[\begin{array}{cccccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -1\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 2\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill -3\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right].$$

   1. Write the solution set of the system in parametric vector form, $\overrightarrow{x}=\overrightarrow{p}+s\overrightarrow{u}+t\overrightarrow{v}$.
   2. Write, without further work, the solution set of the homogeneous system with the same coefficient matrix.
2. Q2The Null Space of a Matrix

   Let $A=\left[\begin{array}{ccccc}1& -1& 2& 0& 3\\ 2& 0& 1& -1& 1\\ 0& 1& -1& 2& 4\end{array}\right]$.

   1. $Nul(A)$ is a subspace of ${ℝ}^k$. What is $k$?
   2. Determine whether $\overrightarrow{u}=(3,-7,-5,1,0)$ and $\overrightarrow{w}=(1,1,1,1,-1)$ belong to $Nul(A)$.
3. Q3The Null Space of a Matrix

   Let $B=\left[\begin{array}{ccccc}1& 2& -1& 0& 3\\ 2& 4& 0& 2& 2\\ -1& -2& 3& 2& -7\end{array}\right]$. Describe $Nul(B)$ as the span of a list of vectors, one for each free variable.

4. Q4The Column Space and the Row Space

   Let $A=\left[\begin{array}{ccc}1& 0& 2\\ 2& 1& 3\\ -1& 1& -4\\ 0& 2& -1\end{array}\right]$. For each of $\overrightarrow{b}=(8,12,-15,-5)$ and $\overrightarrow{c}=(3,5,-4,-1)$, decide whether it lies in $Col(A)$. If it does, write it as a linear combination of the columns of $A$.

5. Q5The Column Space and the Row Space

   Let $A=\left[\begin{array}{ccccc}1& 0& 2& -1& 1\\ 0& 1& 1& 1& -2\\ 1& 1& 4& 0& 0\end{array}\right]$, with rows ${\overrightarrow{r}}_1,{\overrightarrow{r}}_2,{\overrightarrow{r}}_3$.

   1. $Row(A)$ and $Col(A)$ are subspaces of which ${ℝ}^k$?
   2. Show that $Col(A)={ℝ}^3$.
   3. Decide whether $(2,-1,3,-3,4)$ and $(1,1,1,1,1)$ lie in $Row(A)$.
6. Q6Bases for the Null, Column and Row Spaces

   The matrix $A$ below is row equivalent to the matrix $R$ beside it. $$A=\left[\begin{array}{ccccc}1& -3& 0& 1& 3\\ 2& -6& 1& 4& 5\\ -1& 3& 2& 3& -7\\ 3& -9& 1& 5& 8\end{array}\right],R=\left[\begin{array}{ccccc}1& -3& 0& 1& 0\\ 0& 0& 1& 2& 0\\ 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0\end{array}\right].$$ Find a basis for each of $Nul(A)$, $Col(A)$ and $Row(A)$.

7. Q7Bases for the Null, Column and Row Spaces

   Let $W$ be the set of all $(x_1,x_2,x_3,x_4,x_5)\in {ℝ}^5$ satisfying both $$x_1-2x_2+x_3-x_4+3x_5=0\text{and}{x}_2+x_3+2x_4-x_5=0.$$ Write $W$ as the null space of a matrix, and use that to find a basis for $W$ and its dimension.

8. Q8Rank, Nullity and the Rank Theorem

   A matrix $A$ has size $6\times 9$ and $rank(A)=4$. For each of $Nul(A)$, $Col(A)$, $Row(A)$ and $Nul(A^T)$, say which ${ℝ}^k$ it is a subspace of and give its dimension.

9. Q9Rank, Nullity and the Rank Theorem

   Let $A=\left[\begin{array}{cccc}1& 2& 0& 1\\ 2& 5& 1& 1\\ -1& -1& 2& k\\ 1& 1& -2& 1\end{array}\right]$, where $k$ is a real number.

   1. Find $rank(A)$ for every value of $k$.
   2. For the value of $k$ at which the rank is smallest, find a basis of $Nul(A)$ and confirm the rank theorem.
10. Q10Rank and the Consistency of a Linear System

    Let $A=\left[\begin{array}{ccc}1& 2& -1\\ 2& 5& 0\\ 0& 1& 2\\ 3& 7& -1\end{array}\right]$ and $\overrightarrow{b}=(1,3,h,g)$, where $h$ and $g$ are real numbers. Compare $rank(A)$ with the rank of the augmented matrix $[A\mid \overrightarrow{b}]$ to find every pair $(h,g)$ for which $A\overrightarrow{x}=\overrightarrow{b}$ is consistent. For those values, how many free parameters does the solution set have?

11. Q11Synthesis — drawing on several topics in this unit

    Let $$A=\left[\begin{array}{ccccc}1& 0& 2& 0& -1\\ -1& 1& -3& 0& 4\\ 2& 1& 3& 1& 2\\ 0& 1& -1& -1& 2\end{array}\right],\overrightarrow{b}=(1,-1,3,-1).$$

    1. Find $rank(A)$ and a basis of $Nul(A)$, and check the rank theorem.
    2. Verify that $\overrightarrow{p}=(1,0,0,1,0)$ solves $A\overrightarrow{x}=\overrightarrow{b}$, and write the full solution set in parametric vector form.
    3. Is $A\overrightarrow{x}=\overrightarrow{c}$ consistent for every $\overrightarrow{c}\in {ℝ}^4$? If not, give one $\overrightarrow{c}$ for which it is not.

The 8 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

The rest of this unit

The worksheet is the practice, and it is free. Two more printable documents cover the same unit and come with the University Linear Algebra bundle: read the notes first, work this sheet, then sit the test closed-book. See what each one covers.

What the Rank and the Fundamental Subspaces notes cover  About the Rank and the Fundamental Subspaces unit test

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

Parametric vector form

Q1 gives a reduced augmented matrix and asks for the solution set as one vector plus a combination of others. Solve each pivot variable in terms of the free ones, write the whole solution vector, and split it into a constant part and one part per free variable.

The null space

The null space of A is the set of solutions of Ax = 0. Q2 asks two things that need no row reduction at all. Part (a) is about sizes: x must have as many entries as A has columns. Part (b) is a membership test, and membership in a null space is checked by multiplying — compute Au and Aw and see whether each is the zero vector.

Q3 asks for the null space as a span. That is Q1's parametric form with no constant part: reduce B fully, write the general solution, and read off one vector per free variable. Those vectors are automatically independent, so they are also a basis.

The column space and the row space

The column space is the span of the columns — the set of all b for which Ax = b has a solution. Q4 is the membership test from the span set, now named: augment A with each target, reduce, and check consistency. When the answer is yes, the solution gives the weights on the columns.

Q5 contrasts the two spaces. Part (a) is about sizes again, and the two answers differ when A is not square: rows have as many entries as there are columns, and columns as many as there are rows. Part (b) asks you to show the column space is all of three-space; think about how many pivots that needs, and in which rows. Part (c) is membership in the row space, which is the span of the rows.

Bases for all three from one echelon form

Q6 gives A and its reduced form R together and asks for a basis of each space. One reduced matrix answers all three, but not in the same way.

Q7 describes a subspace by two equations and asks you to see it as a null space. The two equations are the rows of a 2 × 5 matrix; the subspace is its null space; Q3's method gives the basis, and the number of free variables gives the dimension.

Rank, nullity and the rank theorem

The rank of A is the dimension of its column space, which is also the dimension of its row space — both equal the number of pivots. The nullity is the dimension of the null space, the number of free variables. Every column is either a pivot column or a free one, which is the whole proof of the theorem.

Q8 asks for the ambient space and the dimension of four subspaces of a matrix you are only told the size and rank of. Everything follows from the theorem above, applied to A and to its transpose, plus the fact that A and its transpose have the same rank.

Q9 has a parameter k in one entry. Row reduce with k in place, and watch for the step where the next pivot depends on k: that entry either is or is not zero, and the two cases give different ranks. Part (b) uses the exceptional value, finds the null space, and checks that rank plus nullity gives the number of columns.

Rank decides consistency

Q10 asks for every pair of parameter values that makes a system consistent. A system Ax = b is consistent exactly when adding b as an extra column does not raise the rank — when no row of zeros in A meets a non-zero entry in b.

Reduce the augmented matrix with h and g in the last column. The rows whose coefficient part becomes zero give conditions on h and g that must hold; those conditions are the answer. The second part is the rank theorem again: the number of free parameters is the number of columns minus the rank.

The synthesis question

Q11 puts the set together on one 4 × 5 matrix. Part (a) is Q6 and Q8: reduce, count pivots, find the null space, and check the theorem. Part (b) is Q1 read backwards — you are given one particular solution, so the full solution set is that vector plus the null space you already have. Part (c) asks whether every right-hand side works, which is a question about the rank compared with the number of rows; if the answer is no, the reduction tells you how to build a right-hand side that fails.

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Getting the most out of it

Write the sizes first

For every matrix, write m × n and say which space each subspace lives in before you compute. Null space and row space live in the space with one coordinate per column; the column space lives in the space with one coordinate per row. Most mix-ups start here.

Circle the pivots

Rank, nullity, the basis of the column space, and consistency all come from the pivot positions. Mark them on the echelon form before reading anything else from it.

Say which matrix each basis came from

Label your column-space basis "pivot columns of A" and your row-space basis "non-zero rows of R". The label is a check on yourself, and a marker looks for it.

Check the rank theorem every time

After finding a rank and a null-space basis, add them. If they do not give the number of columns, there is an error somewhere, and it is cheaper to find it now.

Common questions