Question

Let \( W_1 \) be the real vector space of all \( 5 \times 2 \) matrices such that the sum of the entries in each row is zero. Let \( W_2 \) be the real vector space of all \( 5 \times 2 \) matrices such that the sum of the entries in each column is zero. Then the dimension of the space \( W_1 \cap W_2 \) is .................

Hint:

The dimension of the intersection of two subspaces is the number of independent conditions that both subspaces satisfy.

Answer 1

Correct Answer: 4

1. Dimension of the Universal Space

The dimension of the space of all $5 \times 2$ matrices is the number of independent entries:

$$\dim(M_{5 \times 2}(\mathbb{R})) = 5 \times 2 = 10$$

2. Dimension of $W_1$

$W_1$ is the set of $5 \times 2$ matrices such that the sum of the entries in each row is zero.

For a matrix $A$:

$$A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \vdots & \vdots \\ a_{51} & a_{52} \end{pmatrix}$$

The constraints are:

$$\begin{aligned} a_{11} + a_{12} &= 0 \\ a_{21} + a_{22} &= 0 \\ &\vdots \\ a_{51} + a_{52} &= 0 \end{aligned}$$

There are $\mathbf{5}$ independent linear constraints (one for each row). The total number of variables is 10.

$$\dim(W_1) = \dim(M_{5 \times 2}) - (\text{Number of independent constraints}) = 10 - 5 = \mathbf{5}$$

3. Dimension of $W_2$

$W_2$ is the set of $5 \times 2$ matrices such that the sum of the entries in each column is zero.

The constraints are:

$$\begin{aligned} a_{11} + a_{21} + a_{31} + a_{41} + a_{51} &= 0 \quad (\text{1st column sum}) \\ a_{12} + a_{22} + a_{32} + a_{42} + a_{52} &= 0 \quad (\text{2nd column sum}) \end{aligned}$$

There are $\mathbf{2}$ independent linear constraints (one for each column).

$$\dim(W_2) = 10 - 2 = \mathbf{8}$$

4. Dimension of $W_1 \cap W_2$

$W_1 \cap W_2$ is the set of matrices that satisfy all $5 + 2 = 7$ constraints simultaneously.

We must ensure the 7 constraints are linearly independent.

5 constraints from $W_1$: Row sums are zero.

2 constraints from $W_2$: Column sums are zero.

If we know all the $a_{ij}$ except for the last column, we can determine the last column's entries from the row sum constraints: $a_{i2} = -a_{i1}$ for $i=1$ to 5.

Then, we apply the column sum constraints: $\sum a_{i1} = 0$ and $\sum a_{i2} = 0$.

Since $a_{i2} = -a_{i1}$, the second column sum constraint becomes $\sum (-a_{i1}) = - \sum a_{i1} = 0$.

This means the two column sum constraints are dependent if the row sum constraints are satisfied. If the first 6 constraints are met, the 7th constraint (second column sum) is automatically satisfied.

The set of independent constraints for $W_1 \cap W_2$ is:

Row $i$ sum $= 0$ (5 constraints)

Column 1 sum $= 0$ (1 constraint)

Total number of independent constraints is $\mathbf{6}$.

$$\dim(W_1 \cap W_2) = \dim(M_{5 \times 2}) - (\text{Total independent constraints})$$

$$\dim(W_1 \cap W_2) = 10 - 6 = \mathbf{4}$$