Linear Algebra and Its Applications, Exercise 2.1.4

Exercise 2.1.4. Consider the set of all 3 by 3 symmetric matrices and the set of all 3 by 3 lower triangular matrices, each of which are subspaces of the space of all 3 by 3 matrices. What is the smallest subspace of the space of 3 by 3 matrices that contains both of those subspaces? What is the largest subspace contained in both of those subspaces?

Answer: Any subspace must be closed under vector addition and scalar multiplication. So if a subspace of the set of 3 by 3 matrices contains both the subspace of 3 by 3 symmetric matrices and the subspace of 3 by 3 lower triangular matrices, then it has to also contain all possible linear combinations of those two subspaces. (Otherwise it wouldn’t be a vector space.)

In particular, the subspace we’re looking for has to contain all matrices of the form $S+L$ where $S$ is a 3 by 3 symmetric matrix and $L$ is a 3 by 3 lower triangular matrix. Now, in general symmetric matrices can have nonzero entries above the diagonal, to match corresponding nonzero entries below the diagonal. Therefore in general the sum $S+L$ will produce a matrix that has entries both above and below the diagonal.

The question then becomes, are there any restrictions on what $S+L$ can look like? Or can $S+L$ end up being any possible 3 by 3 matrix? To find out, we take a random 3 by 3 matrix

$A = \begin{bmatrix} a_{11}&a_{12}&a_{13} \\ a_{21}&a_{22}&a_{23} \\ a_{31}&a_{32}&a_{33} \end{bmatrix}$

and try to represent it as the sum of a 3 by 3 symmetric matrix $S$ and a 3 by 3 lower triangular matrix $L$ . Since $L$ must have zeros above the diagonal (by definition), the values for $a_{12}$ , $a_{13}$ , and $a_{23}$ (the entries of $A$ above the diagonal) have to come from $S$ . We can also take the diagonal entries $a_{11}$ , $a_{22}$ , and $a_{33}$ from $S$ as well, since we can just set the diagonal entries of $L$ to be zero. (It will still be lower triangular in this case.)

Since $S$ is a symmetric matrix the entries below the diagonal must be the same as the entries above the diagonal, so we have

$S = \begin{bmatrix} a_{11}&a_{12}&a_{13} \\ a_{12}&a_{22}&a_{23} \\ a_{13}&a_{23}&a_{33} \end{bmatrix}$

What then should $L$ look like? We know the entries above the diagonal are zero (because $L$ is lower triangular) and also that the diagonal entries of $L$ are zero as well (from the previous paragraph). So we have

$L = \begin{bmatrix} 0&0&0 \\ l_{21}&0&0 \\ l_{31}&l_{32}&0 \end{bmatrix}$

and we have to find suitable values for $l_{21}$ , $l_{31}$ , and $l_{32}$ .

For $A$ to equal the sum of $S$ and $L$ we must have

$\begin{array}{rcrcr} a_{21}&=&a_{12}&+&l_{21} \\ a_{31}&=&a_{13}&+&l_{31} \\ a_{32}&=&a_{23}&+&l_{32} \end{array}$

where the first term in each sum above comes from $S$ and the second term from $L$ . Solving for $l_{21}$ , $l_{31}$ , and $l_{32}$ we have

$\begin{array}{rcrcr} l_{21}&=&a_{21}&-&a_{12} \\ l_{31}&=&a_{31}&-&a_{13} \\ l_{32}&=&a_{32}&-&a_{23} \end{array}$

so that

$L = \begin{bmatrix} 0&0&0 \\ a_{21} - a_{12}&0&0 \\ a_{31} - a_{13}&a_{32} - a_{23}&0 \end{bmatrix}$

We then have

$S+L = \begin{bmatrix} a_{11}&a_{12}&a_{13} \\ a_{12}&a_{22}&a_{23} \\ a_{13}&a_{23}&a_{33} \end{bmatrix} + \begin{bmatrix} 0&0&0 \\ a_{21}-a_{12}&0&0 \\ a_{31}-a_{13}&a_{32}-a_{23}&0 \end{bmatrix}$

$= \begin{bmatrix} a_{11}+0&a_{12}+0&a_{13}+0 \\ a_{12}+a_{21}-a_{12}&a_{22}+0&a_{23}+0 \\ a_{13}+a_{31}-a_{13}&a_{23}+a_{32}-a_{23}&a_{33}+0 \end{bmatrix}$

$= \begin{bmatrix} a_{11}&a_{12}&a_{13} \\ a_{21}&a_{22}&a_{23} \\ a_{31}&a_{32}&a_{33} \end{bmatrix} = A$

So, for any 3 by 3 matrix $A$ we can find a 3 by 3 symmetric matrix $S$ and a 3 by 3 lower triangular matrix $L$ for which $S+L = A$ . Put another way, adding 3 by 3 symmetric matrices and 3 by 3 lower triangular matrices together can produce any possible 3 by 3 matrix. So the smallest possible subspace containing both all 3 x 3 symmetric matrices and all 3 by 3 lower triangular matrices is the space of all 3 by 3 matrices.

What about the largest subspace contained in both the subspace of 3 by 3 symmetric matrices and 3 by 3 lower triangular matrices? If a matrix $A$ is in the subspace of lower triangular matrices then we must have $a_{ij} = 0$ for $j > i$ . If $A$ is also in the subspace of symmetric matrices then we must also have $a_{ij} = a_{ji}$ for all $i$ and $j$ , and thus $a_{ij} = 0$ for $j < i$ . So we have $a_{ij} = 0$ for $i \ne j$ and $A$ is a diagonal matrix.

The sum of two diagonal matrices is itself a diagonal matrix, so the set of all diagonal matrices is closed under vector addition. If $D$ is a diagonal matrix and $c$ is a scalar then $cD$ is also a diagonal matrix (even if $c = 0$ ), so the set of all diagonal matrices is closed under scalar multiplication. The set of all 3 by 3 diagonal matrices is therefore a subspace of the space of all 3 by 3 matrices, and it is the largest subspace that is contained in both the subspace of 3 by 3 symmetric matrices and the subspace of 3 by 3 lower triangular matrices.

UPDATE: I completely rewrote the answer to the first part of the problem to provide a more complete explanation, in response to a question from a reader. (I also corrected a typo in the definition of $L$ .)

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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11 Responses to Linear Algebra and Its Applications, Exercise 2.1.4

PA says:

September 19, 2014 at 11:29 am

Isn’t the smallest subspace the zero matrix and the largest the 3*3 matrix?

- hecker says:
  
  September 19, 2014 at 2:38 pm
  
  If we are considering the vector space of all 3 by 3 matrices, and all possible subspaces within that vector space, then, yes, the zero matrix is the smallest possible subspace. However that’s not what the question was asking. It was asking for the smallest subspace that contained both the set of all lower triangular matrices and the set of all symmetric matrices. While the zero matrix is in fact both a lower triangular matrix and a symmetric matrix, there are also nonzero lower triangular matrices and nonzero symmetric matrices. So the zero matrix can’t be the subspace we’re looking for.
  
  As for the other part of your question, the set of all 3 by 3 matrices is the largest possible subspace in the vector space of 3 by 3 matrices. But again that’s not what the question was asking for. It was asking for the largest possible subspace that was contained in the set of all lower triangular matrices and the set of all symmetric matrices. In other words, every matrix in the subspace had to be both lower triangular and symmetric.
  
- PA says:
  
  June 16, 2016 at 12:38 pm
  
  Thank you so much for the detailed reply and sorry about replying so late. Also thank you so much for all the detailed posts on linear algebra.
  
mangoman says:

September 21, 2015 at 6:06 am

isn’t it the SMALLEST subspace DIAGONAL MATRIX? i’m confused

- hecker says:
  
  September 21, 2015 at 2:35 pm
  
  Please read my comment above: The second part of the question is asking for a subspace each of whose members a) is a lower triangular 3 by 3 matrix and b) is a symmetric 3 by 3 matrix. There are multiple subspaces that meet these two conditions, including the subspace of all 3 by 3 diagonal matrices. For example, the subspace containing only the 3 by 3 zero matrix meets this condition, since that matrix is both lower triangular and symmetric. Another example is the subspace of all 3 by 3 matrices that have zeros in every position except the (1, 1) position: All those matrices are again both lower triangular and symmetric, and any linear combination of two such matrices also has zeros in every position except the (1, 1) position. (In other words, the set of such matrices forms a subspace.)
  
  Now, the subspace containing all 3 by 3 matrices with zeros in every position except the (1, 1) position contains the 3 by 3 zero matrix as one member, but not the only member, so that subspace is larger than the subspace consisting only of the zero matrix. Similarly the subspace of all 3 by 3 diagonal matrices contains as elements all 3 by 3 matrices with zeros in every position except the (1, 1) position, but not just those matrices, so the subspace of 3 by 3 diagonal matrices is larger than the subspace of 3 by 3 matrices with zeros in every position except the (1, 1) position.
  
  Is there another subspace of 3 by 3 matrices that is larger than the subspace of 3 by 3 diagonal matrices, and all of whose members are both lower triangular and symmetric? The answer is no. The only way to make a subspace larger than the subspace of 3 by 3 diagonal matrices is to put a nonzero matrix element off the diagonal. But such a matrix can be either lower triangular or symmetric, but not both at the same time. So the subspace of 3 by 3 diagonal matrix is the largest subspace simultaneously meeting the two conditions.
  
  - mangoman says:
    
    September 22, 2015 at 12:48 am
    
    now i got it, thank you so much.
Sam says:

September 7, 2017 at 2:33 pm

Thank for the explanation. But I’m still a bit confused. For finding the largest subspace, the matrix A should fit the requirement of both 3×3 symmetric and lower triangular matrices, so A must be a diagonal matrix. But why we could totally ignore it when doing the first part of the question (finding the smallest subspace) ?

- Sam says:
  
  September 7, 2017 at 2:38 pm
  
  Sorry I misread the question. The first question is finding subspace that contains both subspaces. While the second question is finding subspace that “is contained in” both subspaces. Thanks for your explanation.
  
  - hecker says:
    
    September 7, 2017 at 6:16 pm
    
    No problem, these are somewhat confusing questions.
jamesteow says:

October 13, 2017 at 4:05 am

Thank you very much for elaborating on your answer. This was extremely helpful.