QuestionNYC 05 VectorSpaces: Problem 12 (1 point) (a) If $S$ is the subspace of $M_{5}(\mathbb{R})$ consisting of all upper triangular matrices, then $\operatorname{dim} S=$ $\square$ (b) If $S$ is the subspace of $M_{4}(\mathbb{R})$ consisting of all matrices with trace 0 , then $\operatorname{dim} S=$ $\square$
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Studdy Solution

STEP 1

1. $M_n(\mathbb{R})$ denotes the space of all $n \times n$ matrices with real entries.
2. The dimension of a subspace is the number of vectors in a basis for that subspace.
3. An upper triangular matrix has all zeros below the main diagonal.
4. The trace of a matrix is the sum of its diagonal elements.

STEP 2

1. Determine the dimension of the subspace of upper triangular matrices in $M_5(\mathbb{R})$ .
2. Determine the dimension of the subspace of matrices with trace 0 in $M_4(\mathbb{R})$ .

STEP 3

To find the dimension of the subspace $S$ of upper triangular matrices in $M_5(\mathbb{R})$ , count the number of free entries in an upper triangular matrix.
An upper triangular matrix in $M_5(\mathbb{R})$ has the form: $\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} & a_{15} \\ 0 & a_{22} & a_{23} & a_{24} & a_{25} \\ 0 & 0 & a_{33} & a_{34} & a_{35} \\ 0 & 0 & 0 & a_{44} & a_{45} \\ 0 & 0 & 0 & 0 & a_{55} \\ \end{bmatrix}$
There are $5 + 4 + 3 + 2 + 1 = 15$ free entries (the diagonal and above).

STEP 4

To find the dimension of the subspace $S$ of matrices with trace 0 in $M_4(\mathbb{R})$ , consider the condition that the sum of the diagonal elements is zero.
A general $4 \times 4$ matrix has 16 entries. The trace condition imposes one linear constraint (i.e., $a_{11} + a_{22} + a_{33} + a_{44} = 0$ ), reducing the number of free entries by 1.
Thus, the dimension is $16 - 1 = 15$ .
The dimensions are: (a) $\operatorname{dim} S = 15$ (b) $\operatorname{dim} S = 15$

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