# Rank-Deficiency in an Augmented Matrix comprised of Two Full Rank Matrices

I have a $\boldsymbol{M}_{6\times 6}$ matrix structured as follows:

$$\boldsymbol{A} = \begin{bmatrix}a_1 & a_2 & a_3 & 0 & 0 & 0\\0 & a_1 & 0 & a_2 & a_3 & 0\\0 & 0 & a_1 & 0 & a_2 & a_3\end{bmatrix}$$

$$\boldsymbol{B} = \begin{bmatrix}b_1 & b_2 & b_3 & 0 & 0 & 0\\0 & b_1 & 0 & b_2 & b_3 & 0\\0 & 0 & b_1 & 0 & b_2 & b_3\end{bmatrix}$$

$$\boldsymbol{M} = \begin{bmatrix}\boldsymbol{A}\\\boldsymbol{B}\end{bmatrix} = \begin{bmatrix}a_1 & a_2 & a_3 & 0 & 0 & 0\\0 & a_1 & 0 & a_2 & a_3 & 0\\0 & 0 & a_1 & 0 & a_2 & a_3\\b_1 & b_2 & b_3 & 0 & 0 & 0\\0 & b_1 & 0 & b_2 & b_3 & 0\\0 & 0 & b_1 & 0 & b_2 & b_3\end{bmatrix}$$

I populated $\boldsymbol{A}$, $\boldsymbol{B}$ with several different combinations of values for $a_i, b_i$, e.g., $a_i$ = $[1, 2, 3]$, $b_i$ = $[4, 5, 6]$, to find the rank of $\boldsymbol{M}$ using MATLAB.

It turns out that $Rank(\boldsymbol{A})=3$ and $Rank(\boldsymbol{B})=3$, but, $\boldsymbol{M}$ is rank-deficient with $Rank(\boldsymbol{M})=5$. I was not able to find the dependent row in $\boldsymbol{M}$. All rows look like linearly independent.

I tried to find the rank of M([r1, r2, r3, r4, r5],:) but no matter what combination of 5 rows I pick, the rank is 5 which implies that any 5 rows combinations are linearly independent. Does this mean the 6th row is a combination of other 5 rows?

Could someone kindly help me how I can find the row in $\boldsymbol{M}$ that is linearly dependent and makes $\boldsymbol{M}$ rank deficient?

## 2 Answers

I tried to find the rank of M([r1, r2, r3, r4, r5],:) but no matter what combination of 5 rows I pick, the rank is 5 which implies that any 5 rows combinations are linearly independent. Does this mean the 6th row is a combination of other 5 rows?

Not quite. It means that there are linear combination of the six rows (or the six columns) that sum to zero. Therefore, any one of the rows (or columns) can be expressed as a linear combination of the other 5 rows (or 5 columns).

Could someone kindly help me how I can find the row in M that is linearly dependent and makes M rank deficient?

As stated above, any one of the columns can be expressed as a nontrivial linear combination of the other 5 columns. ("Nontrivial" i.e. the coefficients of the linear combination are not all zero.) I will show how to find such a linear combination.

The reduced row echelon form $M$ is

$$\textrm{rref}(M) = \left[ \begin {array}{cccccc} 1&0&0&0&0&-{\dfrac { \left( {a_2}{ b_3}-{a_3}{b_2} \right) ^{2}}{ \left( {b_2}{a_1}-{ b_1}{a_2} \right) ^{2}}}\\ 0&1&0&0&0&{\dfrac { \left( {b_3}{a_1}-{b_1}{a_3} \right) \left( {a_2 }{b_3}-{a_3}{b_2} \right) }{ \left( {b_2}{a_1}-{ b_1}{a_2} \right) ^{2}}}\\ 0&0&1&0&0&-{ \dfrac {{a_2}{b_3}-{a_3}{b_2}}{{b_2}{a_1}-{ b_1}{a_2}}}\\ 0&0&0&1&0&-{\dfrac { \left( {b_3 }{a_1}-{b_1}{a_3} \right) ^{2}}{ \left( {b_2}{a_1 }-{b_1}{a_2} \right) ^{2}}}\\ 0&0&0&0&1&{ \dfrac {{b_3}{a_1}-{b_1}{a_3}}{{b_2}{a_1}-{ b_1}{a_2}}}\\ 0&0&0&0&0&0\end {array} \right]$$

(I computed this using the Maple software.) The fact that the last row is all zero implies that the $M$ is rank-deficient.

Moreover, the $\textrm{rref}(M)$ implies that

$$\vec{x} = \left[ \begin {array}{c} {\dfrac { \left( {a_2}{b_3}-{a_3} {b_2} \right) ^{2}s}{ \left( {b_2}{a_1}-{b_1}{a_2 } \right) ^{2}}}\\ -{\dfrac { \left( {b_3}{ a_1}-{b_1}{a_3} \right) \left( {a_2}{b_3}-{a_3}{ b_2} \right) s}{ \left( {b_2}{a_1}-{b_1}{a_2} \right) ^{2}}}\\ {\dfrac { \left( {a_2}{b_3} -{a_3}{b_2} \right) s}{{b_2}{a_1}-{b_1}{a_2}}} \\ {\dfrac { \left( {b_3}{a_1}-{b_1}{ a_3} \right) ^{2}s}{ \left( {b_2}{a_1}-{b_1}{a_2} \right) ^{2}}}\\ -{\dfrac { \left( {b_3}{a_1 }-{b_1}{a_3} \right) s}{{b_2}{a_1}-{b_1}{a_2}} }\\ s\end {array} \right]$$

spans the null space of $M$: in other words, for any real number $s$, $$M \vec{x} = \vec{0}. \tag{*}$$

Equation (*) is equivalent with the following equation

$$\sum_{n=1}^6 x_n\vec{M}_n = 0, \tag{\dagger}$$

where "$\vec{M}_n$" denotes the $n$th column of the matrix $M$. Since $x_n$ is a function of $s$, simply substitute one nonzero value of $s$. Then, Equation ($\dagger$) indicates a nontrivial linear combination of the column vectors of $M$ that sums to zero.

• I was just in the middle of doing the same thing! Try looking at the RREF of the transpose, for a much nicer linear combination of rows. Aug 1, 2018 at 3:08
• I didn't think to try that - the RREF of $A^\top$ does look much nicer. Thank you @TheoBendit!
– Kyle
Aug 1, 2018 at 3:10

I was hoping Kyle might mention it, but

$$b_1r_1 + b_2 r_2 + b_3 r_3 = a_1r_4 + a_2 r_5 + a_3 r_6,$$

which gives us our linear dependency.

• Thank you. Simple and to the point. Aug 1, 2018 at 6:57
• I wish I could mark two answers to my question. Aug 1, 2018 at 18:24
• Keep Kyle's. He shows more or less how I got to this conclusion. :-) Aug 1, 2018 at 23:57