LINEAR-ALGEBRA · Unit 11 · Video 1 · Interactive Practice

Beyond Column Vectors: The Dimension Formula Hiding Inside Matrix Spaces

IKey Formulas

Formula	Name	Description
$\dim(M_{n \times n}) = n^2$	Full matrix space	All $n \times n$ real matrices
$\dim(\mathrm{Sym}_n) = \frac{n(n+1)}{2}$	Symmetric matrices	$A^T = A$ ; free entries on & above diagonal
$\dim(\mathrm{UT}_n) = \frac{n(n+1)}{2}$	Upper triangular	Zeros forced below diagonal
$\dim(V) + \dim(W) = \dim(V \cap W) + \dim(V + W)$	Dimension formula	For any two subspaces $V, W$

IIVisualization 1 — Counting Free Entries

Each constraint on a matrix's entries removes a free choice; count the free entries and you have the dimension.

Subspace Size n 3

💡 Challenge: Symmetric and Upper Triangular share the same dimension — which subspace is their intersection $S \cap U$ ?

IIIVisualization 2 — Decomposing $A = S + U$

Every square matrix splits uniquely into a symmetric part $S$ plus a strictly upper-triangular part $U$ .

Load A

💡 Challenge: Edit $A$ 's entries until $U$ becomes the zero matrix — what kind of matrix must $A$ be?

IVVisualization 3 — The Dimension Formula

Like inclusion–exclusion for sets, adding two subspaces' dimensions double-counts their overlap — the formula rebalances it.

Size n 3

VQuiz Questions

Question 1

What is the dimension of the vector space of all $4 \times 4$ symmetric matrices?

Hint: Count the entries you can freely choose — those on and above the main diagonal.

✅ Correct! There are 4 diagonal + 6 above-diagonal = 10 free entries.

❌ Not quite. Only count entries on and above the diagonal — below-diagonal entries are forced copies due to symmetry.

Show solution

Solution:

For a $4 \times 4$ symmetric matrix, the free entries are on and above the diagonal:

Diagonal: 4 entries $(a_{11}, a_{22}, a_{33}, a_{44})$
Above diagonal: $\binom{4}{2} = 6$ entries $(a_{12}, a_{13}, a_{14}, a_{23}, a_{24}, a_{34})$
Below diagonal: forced by symmetry $(a_{ji} = a_{ij})$

$\dim(\mathrm{Sym}_4) = \frac{4 \times 5}{2} = \frac{20}{2} = 10$

Question 2

True or False: The union $S \cup U$ of two subspaces is always a subspace.

✅ Correct! The union fails closure under addition. The sum of subspaces $S + U$ is the right construction.

❌ Not quite. Consider adding a symmetric matrix to an upper triangular one — must the result lie in the union?

Show solution

Answer: False

The union of two subspaces is almost never a subspace (unless one contains the other).

Counterexample from the video: Take $s \in S$ (symmetric) and $u \in U$ (upper triangular):

$s = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \quad u = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$

Both are in $S \cup U$ . But their sum:

$s + u = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$

is neither symmetric (entry $(2,3) = 1 \neq 0 =$ entry $(3,2)$ ) nor upper triangular (entry $(2,1) = 1 \neq 0$ ).

So $s + u \notin S \cup U$ — the union is not closed under addition!

The correct construction is the sum of subspaces: $S + U = \{s + u : s \in S,\; u \in U\}$ , which is always a subspace.

Question 3

Suppose $V$ and $W$ are subspaces of a vector space with $\dim(V) = 8$ , $\dim(W) = 5$ , and $\dim(V \cap W) = 2$ .

What is $\dim(V + W)$ ?

✅ Correct! $\dim(V + W) = 8 + 5 - 2 = 11$ . The overlap gets subtracted, not double-counted.

❌ Not quite.

Show solution

Solution:

Apply the dimension formula:

$\dim(V) + \dim(W) = \dim(V \cap W) + \dim(V + W)$

$8 + 5 = 2 + \dim(V + W)$

$\dim(V + W) = 13 - 2 = 11$

The 2-dimensional intersection was counted once in $\dim(V)$ and once in $\dim(W)$ , so we subtract it — exactly like inclusion-exclusion.

Question 4

Which of the following is a subspace of the vector space of all $3 \times 3$ matrices?

Remember: a subspace must contain the zero matrix and be closed under addition and scalar multiplication.

✅ Correct! Diagonal matrices form a 3-dimensional subspace — closed under addition and scalar multiplication, containing the zero matrix.

❌ Not quite.

Show solution

Answer: All 3×3 diagonal matrices

Check each option against the subspace axioms:

det = 0: NOT a subspace. Take $A = \mathrm{diag}(1,0,0)$ and $B = \mathrm{diag}(0,1,1)$ : both have det $= 0$ , but $A + B = I$ has det $= 1 \neq 0$ . Not closed under addition.
✅ Diagonal matrices: IS a subspace.
- Sum of two diagonal matrices is diagonal ✓
- Scalar times a diagonal matrix is diagonal ✓
- The zero matrix is diagonal ✓
- $\dim = 3$
All entries positive: NOT a subspace. The zero matrix has no positive entries (excluded). Also, $(-1) \cdot A$ has all negative entries.
Invertible matrices: NOT a subspace. The zero matrix is not invertible (excluded). Also, $A + (-A) = 0$ is not invertible.

Solved: 0 / 4

Beyond Column Vectors: The Dimension Formula Hiding Inside Matrix Spaces

IKey Formulas

IIVisualization 1 — Counting Free Entries

IIIVisualization 2 — Decomposing A=S+UA = S + UA=S+U

IVVisualization 3 — The Dimension Formula

VQuiz Questions

IIIVisualization 2 — Decomposing $A = S + U$