I have the following theorem:
Theorem 3.12. Let A be an m n matrix. Then the dimension of its row
space is equal to the dimension of its column space.
And the following proof is given:
Proof. Suppose that $\lbrace v_1,v_2,\dots,v_k\rbrace$ is a basis for the column space of $A$. Then each column of $A$ can be expressed as a linear combination of these vectors; suppose that the $i$-th column $c_i$ is given by $$c_i = \gamma_{1i}v_1+\gamma_{2i}v_2+\dots+\gamma_{ki}v_k$$ Now form two matrices as follows: $B$ is an $m\times k$ matrix whose columns are the basis vectors $v_i$, while $C=(\gamma_{ij})$ is a $k\times n$ matrix whose $i$-th column contains the coefficients $\gamma_{1i},\gamma_{2i,}\dots,\gamma_{ki}$. It then follows$^7$ that $A=BC$.
However, we can also view the product $A= BC$ as expressing the rows of $A$ as a linear combination of the rows of $C$ with the $i$-th row of $B$ giving the coefficients for the linear combination that determines the $i$-th row of $A$. Therefore, the rows of $C$ are a spanning set for the row space of $A$, and so the dimension of the row space of $A$ is at most $k$. We conclude that: $$\dim(\operatorname{rowsp}(A))\leq\dim(\operatorname{colsp}(A))$$
Applying the same argument to $A^t$, we conclude that:$$\dim(\operatorname{colsp}(A))\leq\dim(\operatorname{rowsp}(A))$$and hence these values are equal
However, I am finding this proof impossible to follow and understand. Can someone please offer an alternative proof or explain what this proof is saying?
Thank you.
Best Answer
You can consider it as the next explanation also for the fact that the row dimension of a Matrix equals the column dimension of a matrix. For that I will use what it's called the rank of a Matrix.
The rank $r$ of a Matrix can be defines as the number of non-zero singular values of the Matrix, So applying the singular value decomposition of the matrix, we get $A=U\Sigma V^T$. This implies that the range $dim(R(A))=r$, as the range of $A$ is spanned by the first $r$ columns of $U$. We know that the range of $A$ is defined as the subspace spanned by the columns of $A$, so the dimension of it will be $r$.
If we take the transpose of the Matrix and compute it's SVD, we see that $A^T=V\Sigma^T U^T$, and as the Sigma Matrix remains the same number of non-zero elements as the one for $A$, the rank of this Matrix will still be $r$. So as done for $A$, the dimension for the range of $A^T$ is equal to $r$ too, but as the range of $A^T$ is the row space of $A$, we conclude that the dimension for both spaces must be the same and equal to the range of the Matrix $A$.