Question

Consider the following statements for the linear transformation \(L\) on \(\mathbb{R}^3\) defined by \(L(x,y,z)=(x,y,0)\) for all \((x,y,z)\in\mathbb{R}^3\):

(I) Rank of \(L\) is 2.

(II) The only eigenvalues of \(L\) are 0 and 1.

(III) The number of linearly independent eigenvectors of \(L\) is 2.

Choose the correct answer:

• A. Only (I) and (II) are true
• B. Only (II) and (III) are true
• C. Only (I) and (III) are true
• D. All (I), (II) and (III) are true

Hint:

L is the projection onto the xy-plane; projections satisfy L^2=L, forcing eigenvalues 0 and 1 only.

Answer 1

The Correct Option is A

Step 1: Write \(L\) as a matrix with respect to the standard basis \(\{e_1,e_2,e_3\}\) of \(\mathbb{R}^3\). \[L(x,y,z) = (x,y,0) \implies A = \begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}\]

Step 2: Check statement (I). The image of \(L\) is \(\{(x,y,0): x,y\in\mathbb{R}\}\), spanned by \(e_1\) and \(e_2\), which are linearly independent, so \[\operatorname{rank}(L) = 2\] Statement (I) is true.

Step 3: Check statement (II). Since \(A\) is already diagonal, its eigenvalues are its diagonal entries: 1, 1, 0. So the only eigenvalues of \(L\) are 0 and 1, and statement (II) is true.

Step 4: Check statement (III). For \(\lambda = 1\): solving \((A-I)v=0\) gives \(z=0\) with \(x,y\) free, an eigenspace \(\operatorname{span}\{e_1,e_2\}\) of dimension 2. For \(\lambda = 0\): solving \(Av=0\) gives \(x=0, y=0\) with \(z\) free, an eigenspace \(\operatorname{span}\{e_3\}\) of dimension 1. Together \(e_1, e_2, e_3\) form three linearly independent eigenvectors, since \(A\) is diagonal and hence diagonalizable with a full set of 3 independent eigenvectors. So the count is 3, not 2, and statement (III) is false.

Step 5: Combine the results: (I) and (II) are true, (III) is false.

\[\boxed{\text{Only (I) and (II) are true}}\]