Basis and Dimension Questions and Answers

Answer: D

The dimension is \(\dfrac{n(n - 1)}{2}\).

A skew symmetric matrix satisfies \(a_{ji} = -a_{ij}\). Taking \(i = j\) gives \(a_{ii} = -a_{ii}\), so every diagonal entry is 0. The entries above the diagonal can be chosen freely, and they fix those below. There are \(n(n - 1)/2\) positions above the diagonal, and the matrices \(E_{ij} - E_{ji}\) (\(i \lt j\)) form a basis.

\(n(n + 1)/2\) is the dimension of the symmetric matrices, which also keep the diagonal. \(n^{2} - n\) counts all off-diagonal positions without using the skew condition, and \(n - 1\) is the dimension of the diagonal matrices of trace zero.

Answer: B

The dimension is \(n^{2} - 1\).

The trace is a linear map from the \(n^{2}\)-dimensional space of all \(n \times n\) real matrices onto \(\mathbf{R}\). It is onto (the matrix \(E_{11}\) has trace 1), so its image has dimension 1, and by rank-nullity its kernel, the matrices of trace zero, has dimension \(n^{2} - 1\).

Put differently, the single condition \(a_{11} + a_{22} + \ldots + a_{nn} = 0\) removes one degree of freedom from the \(n^{2}\) entries.

\(n^{2} - n\) would mean the whole diagonal is forced to zero, \(n(n - 1)/2\) is the dimension of the skew symmetric matrices, and \(n - 1\) is the dimension of the diagonal matrices with trace zero.

Answer: C

The real dimension is \(n^{2} + n\).

A symmetric \(n \times n\) matrix is fixed by its entries on and above the diagonal: \(n\) diagonal entries and \(n(n - 1)/2\) above it, so \(n(n + 1)/2\) free entries. Over \(\mathbf{C}\) that is the dimension. Over \(\mathbf{R}\) each complex entry contributes two real parameters (its real and imaginary parts), so the dimension doubles:

\[ 2 \cdot \frac{n(n + 1)}{2} = n^{2} + n \]

\((n^{2} + n)/2\) is the complex dimension, \(n^{2}\) is the dimension of all real (or complex) matrices, and \(n^{2} - n\) has no such meaning here.

Answer: B

The quotient \(\mathbf{R}^{4}/W\) has dimension 2.

Check the spanning vectors for dependence:

\[ [1\ 2\ 0\ 0] - [0\ 1\ 2\ 0] = [1\ 1\ -2\ 0] \]

So the third vector is the difference of the first two, and the first two are clearly independent. Hence dim W = 2 and

\[ \dim \mathbf{R}^{4}/W = 4 - 2 = 2 \]

Assuming the three vectors independent gives 1; counting only the zero fourth coordinate gives 3.

Answer: C

The true statement is \(\dim V = \dim W + \dim V/W\).

The quotient map \(V \to V/W\) is onto with kernel W, so by rank-nullity \(\dim V = \dim W + \dim V/W\). Equivalently, extend a basis of W to a basis of V; the added vectors give a basis of V/W.

W/V is not defined, since W is the smaller space. A minus sign, or leaving out \(\dim W\), gives the wrong count; for \(V = \mathbf{R}^{3}\) and W a plane, \(\dim V/W = 1\) and only \(3 = 2 + 1\) works.

Answer: B

The dimension is \(n + 1\).

Every polynomial of degree at most \(n\) is uniquely \(a_{0} + a_{1}x + \cdots + a_{n}x^{n}\), so \(\{1, x, x^{2}, \ldots, x^{n}\}\) is a basis. It has \(n + 1\) elements, because the constant term is counted as well as the powers 1 to \(n\).

The answer \(n\) forgets the constant polynomial 1, \(n^{2}\) has no connection with this space, and \(\infty\) is the dimension of the space of all polynomials, with no bound on the degree.

Answer: C

The dimension is \(\dfrac{n^{2} - n}{2}\).

A skew symmetric matrix has \(a_{ji} = -a_{ij}\), so every diagonal entry is 0 and the entries below the diagonal are fixed by those above. The free entries are the \(a_{ij}\) with \(i \lt j\), and there are \(n(n - 1)/2\) of them. The matrices \(E_{ij} - E_{ji}\) for \(i \lt j\) form a basis.

\(n^{2}\) is the dimension of all \(n \times n\) matrices, \(n^{2} - n\) counts all off-diagonal positions without using the skew symmetry, and \((n^{2} + n)/2\) is the dimension of the symmetric matrices.

Answer: B

\(W^{\perp}\) is generated by \((1, 2, 3, 4)^{t}\) and \((1, -1, 1, 0)^{t}\).

W is the set of vectors orthogonal to \(a = (1, 2, 3, 4)\) and \(b = (1, -1, 1, 0)\), since its two equations say \(a \cdot x = 0\) and \(b \cdot x = 0\). So \(W = \mathrm{span}\{a, b\}^{\perp}\), and in a finite dimensional space \(W^{\perp} = \mathrm{span}\{a, b\}\). The two vectors are independent, so \(\dim W^{\perp} = 2\) and \(\dim W = 2\).

Option A spans a 3-dimensional space, too large. Options C and D give vectors that are not in the span of \(a\) and \(b\); for instance \((0, 0, 0, 1)\) would need a combination of \(a, b\) with zero first three entries, which forces both coefficients to be 0.