0
$\begingroup$

I am currently watching a lecture series on Youtube by Dr. Gilbert Strang, and it's been great so far. The only issue that I'm running into is that, without practicing the problems, I have no way to really gauge if I'm really understanding the material, and the concepts aren't sticking as well.

Specifically, we are currently covering the four major subspaces of linear algebra: the column space, the null space, the row space, and the left null space. One of the ideas in this chapter is to be able to find the basis for each of these spaces given a certain mxn matrix, and determining whether or not columns or rows are linearly independent.

I mostly understand the concepts themselves, but I really need to practice to be sure, and to get it to stick. I've tried searching google for practice problems, as well as past questions on this site, and so far I'm not really finding what I need. Would anyone happen to know a good source of practice problems for subspace bases and linear independence of vectors for matrices at the first year (undergraduate) linear algebra level? Thanks, I would appreciate it.

$\endgroup$
  • 1
    $\begingroup$ Strang has multiple textbooks on linear algebra; I'm sure they will have plenty of problems and probably will align fairly well with his lecture series. $\endgroup$ – Theo Bendit Jul 5 '19 at 0:53
  • $\begingroup$ Thanks! Now that you mention it, looking back at lecture 1 of the online course, he lists the book on his blackboard at the beginning of the class. $\endgroup$ – erf x Jul 5 '19 at 1:03
  • $\begingroup$ Notably, there is his Linear Algebra and Its Applications. I used to have a copy and liked it. Used it for reference. There was an emphasis on applications, as the title suggests. $\endgroup$ – Chris Custer Jul 5 '19 at 1:31
-1
$\begingroup$

Here is a practical way to find the four subspaces you mention. By finding I mean: find a basis for them. (Because if we know a basis for a subspace, then we know that every other vector in the subspace is a linear combination of the basis vectors).

I am assuming you know that every linear function $f:{\mathbb R}^n \longrightarrow {\mathbb R}^m$ is given by an $m\times n$ matrix $A$ in the sense that $f(x)=Ax$ for every $x$ in ${\mathbb R}^n$. So assume we have such a matrix $A$.

For the column space: compute $A^T$ (the transpose), then do row operations until $A^T$ is in Row Echelon Form. Then the non-zero rows of the Row Echelon Form will be a basis for the Column Space (that is the same thing as the range of the linear function).

For the Null Space: put $A$ in Row Echelon Form. Remember that the leading entry of a row is the first non-zero entry of that row (if there is one). In the Row Echelon Form of $A$, each column that has no pivot(=no entry that is a leading entry of some row) corresponds to a free variable. If you find no free variables, the null space is the just the zero space (only has the zero vector in it, which means the function if one-to-one). If you find $k$ free variables, that means the Null Space has dimension $k$ (meaning: there will be $k$ vectors in a basis for it). To find the $k$ vectors for a specific basis, proceed as follows: set one free variable to $1$, and all others to zero, and solve the homogeneous system corresponding to the Row Echelon Form. The answer will be a vector in a basis for the Null Space. Now set another free variable to $1$ and all the others to $0$, and solve the linear system again, to get a second vector in a basis. So on, until you finish all the free variables, and get $k$ vectors for a basis for the null space.

As for the Row Space, it's just like the column space, but you work on $A$ instead of $A^T$.

The left Null Space is the same thing as the Null Space of $A^T$, so again switch $A$ with $A^T$.

To practice, try the three matrices below. Find a basis for the four subspaces in each case.

$A=\left(\begin{array}{ccc} -1 & 1 & 1 \\ 3 & -1 & 0 \\ 2 & -4 & -5 \end{array} \right)$.

$B=\left(\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 5 & 3 \\ 1 & 0 & 8 \end{array} \right)$.

$C=\left(\begin{array}{ccc} -1 & 1 & 2 \\ -2 & 2 & 4 \\ 3 & -3 & -6 \end{array} \right)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.