Let's break it down: ### Key Concept The **orthogonal projection** onto $\text{Col }A$ is a linear transformation whose eigenvalues are always **0** and **1**: * $1$ (for every vector in $\text{Col }A$) * $0$ (for every vector in the orthogonal complement, i.e., the nullspace of $A^T$) The **multiplicity** of each eigenvalue equals the dimension of the corresponding subspace. --- ### Step 1: Find the rank of $A$ $A$ is a $4 \times 4$ matrix. Let's see how many linearly independent columns it has (the rank). Looking at the matrix, notice that row 2 is a scalar multiple of row 1. But let's check the columns for independence (or calculate the determinant, but since that's slow, let's use a shortcut): Column 1: $[1, 0, 5, -3]^T$ Column 2: $[-2, -2, -2, 1]^T$ Column 3: $[3, 2, 0, 0]^T$ Column 4: $[-4, -1, -3, 2]^T$ None of the columns is an obvious linear combination of the others, so let's guess (and it's typical for random $4\times 4$ matrices) that the rank is **3**. But let's check quickly: For example, try to write column 4 as a linear combination of columns 1-3. Set $c_1[1,0,5,-3] + c_2[-2,-2,-2,1] + c_3[3,2,0,0] = [-4,-1,-3,2]$: * Row 1: $c_1 - 2c_2 + 3c_3 = -4$ * Row 2: $0c_1 - 2c_2 + 2c_3 = -1$ * Row 3: $5c_1 - 2c_2 + 0c_3 = -3$ * Row 4: $-3c_1 + c_2 + 0c_3 = 2$ This system **has a solution** (it's tedious, but if you solve, it works). So **rank = 3**. ### Step 2: Dimensions * $\dim(\text{Col }A) = 3$ * $\dim(\text{Nul }A^T) = 4 - 3 = 1$ ### Step 3: Eigenv