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PROBLEM

NYC 05 VectorSpaces: Problem 12
(1 point)
(a) If SS is the subspace of M5(R)M_{5}(\mathbb{R}) consisting of all upper triangular matrices, then dimS=\operatorname{dim} S= \square
(b) If SS is the subspace of M4(R)M_{4}(\mathbb{R}) consisting of all matrices with trace 0 , then dimS=\operatorname{dim} S= \square
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STEP 1

1. Mn(R) M_n(\mathbb{R}) denotes the space of all n×n 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 M5(R) M_5(\mathbb{R}) .
2. Determine the dimension of the subspace of matrices with trace 0 in M4(R) M_4(\mathbb{R}) .

STEP 3

To find the dimension of the subspace S S of upper triangular matrices in M5(R) M_5(\mathbb{R}) , count the number of free entries in an upper triangular matrix.
An upper triangular matrix in M5(R) M_5(\mathbb{R}) has the form:
[a11a12a13a14a150a22a23a24a2500a33a34a35000a44a450000a55]\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 5 + 4 + 3 + 2 + 1 = 15 free entries (the diagonal and above).

SOLUTION

To find the dimension of the subspace S S of matrices with trace 0 in M4(R) M_4(\mathbb{R}) , consider the condition that the sum of the diagonal elements is zero.
A general 4×4 4 \times 4 matrix has 16 entries. The trace condition imposes one linear constraint (i.e., a11+a22+a33+a44=0 a_{11} + a_{22} + a_{33} + a_{44} = 0 ), reducing the number of free entries by 1.
Thus, the dimension is 161=15 16 - 1 = 15 .
The dimensions are:
(a) dimS=15\operatorname{dim} S = 15
(b) dimS=15\operatorname{dim} S = 15

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